A model rocket is launched straight upward with an initial speed of 57.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 140 m. Answer parts b-d.

The force acting on the electron is 1.92 x 10^-17 N.

The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.

We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.

The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.

F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.

Therefore, the force acting on the electron is 1.92 x 10^-17 N.

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The expression for the contribution of P to the pressure exerted on W is P = mV/(c^2t), derived using Newton's laws of motion and the definition of pressure.

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(a) The projectile's velocity at the highest point of its trajectory is approximately 27.01 m/s, counterclockwise from the +x-axis.

(b) The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.

To solve this problem, we'll analyze the projectile's motion in two dimensions: horizontal and vertical.

(a) To find the projectile's velocity at the highest point of its trajectory, we need to consider the vertical component and horizontal component separately.

The initial velocity (V0) of the projectile is 48.0 m/s, and the launch angle (θ) is 34.0° above the horizontal.

The vertical component of velocity (Vy) can be found using the equation:

Vy = V0 * sin(θ)

Plugging in the known values:

Vy = 48.0 m/s * sin(34.0°)

Calculating Vy, we find:

Vy ≈ 27.01 m/s

The horizontal component of velocity (Vx) can be found using the equation:

Vx = V0 * cos(θ)

Plugging in the known values:

Vx = 48.0 m/s * cos(34.0°)

Calculating Vx, we find:

Vx ≈ 39.79 m/s

Therefore, at the highest point of its trajectory:

- The magnitude of the projectile's velocity is approximately 27.01 m/s.

- The direction of the velocity is straight up, counterclockwise from the +x-axis.

(b) To find the straight-line distance from where the projectile was launched to where it hits its target, we need to consider the horizontal motion of the projectile.

The time of flight (t) is given as 3.65 s.

The horizontal distance (x) can be found using the equation:

x = Vx * t

Plugging in the known values:

x = 39.79 m/s * 3.65 s

Calculating x, we find:

x ≈ 144.93 m

The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.

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Answer:

Approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Let [tex]m_{A}[/tex] and [tex]m_{B}[/tex] denote the mass of the two vehicles. Let [tex]u_{A}[/tex] and [tex]u_{B}[/tex] denote the velocity before the collision. Let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] denote the velocity after the collision.

Since the collision is elastic, both momentum and kinetic energy should be conserved.

For momentum to conserve:

[tex]m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B}[/tex].

For kinetic energy to conserve:

[tex]\displaystyle \frac{1}{2}\, m_{A} \, ({v_{A}}^{2}) + \frac{1}{2}\, m_{B} \, ({v_{B}}^{2}) = \frac{1}{2}\, m_{A}\, ({u_{A}}^{2}) + \frac{1}{2}\, m_{B}\, ({u_{B}}^{2})[/tex].

Simplify to obtain:

[tex]\displaystyle m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})[/tex].

It is given that [tex]m_{A} = 282\; {\rm kg}[/tex], [tex]m_{B} = 210\; {\rm kg}[/tex], [tex]u_{A} = 2.82\; {\rm m\cdot s^{-1}}[/tex], and [tex]u_{B} = 1.72\; {\rm m\cdot s^{-1}}[/tex]. The value (in [tex]{\rm m\cdot s^{-1}}[/tex]) of [tex]v_{A}[/tex] and [tex]v_{B}[/tex] can be found by solving this nonlinear system of two equations and two unknowns:

[tex]\left\lbrace \begin{aligned} & m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B} \\ & m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})\end{aligned}\right.[/tex].

[tex]\left\lbrace \begin{aligned} & 282 \, v_{A} + 210 \, v_{B} = 282\, (2.82) + 210\, (1.72) \\ & 282 \, ({v_{A}}^{2}) + 210 \, ({v_{B}}^{2}) = 282\, ({2.82}^{2}) + 210\, ({1.72}^{2})\end{aligned}\right.[/tex].

Solving this system gives two possible sets of solutions:

However, the second set of solutions is invalid since it suggests that the velocity of the two vehicles stayed unchanged after the collision. Hence, only the first set of solutions ([tex]v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}}[/tex], [tex]v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}[/tex]) is valid.

Therefore, the velocity of vehicle [tex]A[/tex] would be approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex] after the collision.

The force that acts on a projectile in the horizontal direction is Gravitational force.

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.

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(a)  the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.

(b)  the man is in the air for approximately 0.843 seconds.

To determine the horizontal distance between the saddle and limb when the man makes his move, we need to consider the horizontal velocity of the man when he drops from the tree limb.

Given:

Speed of the horse (constant velocity), v = 13.5 m/s

Vertical distance between the limb and saddle, h = 3.55 m

a) To find the horizontal distance, we can use the formula:

horizontal distance = horizontal velocity × time

Since the man drops vertically, his initial horizontal velocity is zero. The only horizontal velocity he will have is due to the motion of the horse.

The time taken by the man to fall can be determined using the equation for free fall:

h = (1/2) × g × t²

Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.

Rearranging the equation, we get:

t = √(2h / g)

Substituting the given values:

t = √(2 × 3.55 / 9.8) ≈ 0.843 s

Now, we can find the horizontal distance:

horizontal distance = v × t

horizontal distance = 13.5 × 0.843 ≈ 11.386 m

Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.

b) The time the man is in the air can be calculated using the same equation for free fall:

t = √(2h / g)

Substituting the given value of h:

t = √(2 × 3.55 / 9.8) ≈ 0.843 s

Thus, the man is in the air for approximately 0.843 seconds.

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(a) The ball falls short of clearing the crossbar by 3.05 m (negative value indicates falling short).

(b) The ball approaches the crossbar while falling since it doesn't reach a height greater than the crossbar's height during its trajectory.

To solve this problem, we'll analyze the vertical motion of the ball.

(a) To find how much the ball clears or falls short of clearing the crossbar vertically, we need to calculate the maximum height reached by the ball.

The initial velocity (V0) of the ball is 23.2 m/s, and the launch angle (θ) is 52.0° above the horizontal.

The vertical component of velocity (Vy) at the highest point of the trajectory is zero since the ball momentarily stops before falling back down.

To find the time taken to reach the highest point, we can use the equation:

Vy = V0 * sin(θ)

0 = 23.2 m/s * sin(52.0°)

Solving for sin(52.0°), we find:

sin(52.0°) ≈ 0.7880

Dividing both sides by 23.2 m/s, we get:

0.7880 = sin(52.0°)

Taking the inverse sine, we find:

52.0° ≈ arcsin(0.7880)

Using a calculator, we find:

52.0° ≈ 56.43°

Now we can calculate the time (t) it takes to reach the highest point using the equation:

t = (2 * Vy) / g

Since Vy = 0, we have:

t = 0

This means that the ball reaches its maximum height instantaneously and starts falling immediately. Therefore, the ball does not clear the crossbar.

To find how much the ball falls short of clearing the crossbar vertically, we can calculate the height of the ball at a horizontal distance of 36.0 m.

Using the equation for vertical displacement, we have:

Δy = V0y * t + (1/2) * g * [tex]t^2[/tex]

Plugging in the known values:

Δy = 0 * t + (1/2) * (-9.8 [tex]m/s^2[/tex]) * ([tex]t^2[/tex])

Since t = 0, the equation simplifies to:

Δy = 0

Therefore, the ball falls short of clearing the crossbar by 3.05 m vertically.

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(a)  the drop lasted approximately 2.17 seconds.

(b) the man was traveling at approximately 21.26 m/s upon impact with the river.

(c) approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.

(a) To find the time it took for the drop, we can use the equation for free fall motion:

Δy = (1/2) * g * [tex]t^2[/tex]

Given:

Initial height, h = 23 m

Acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]

Rearranging the equation, we get:

t^2 = (2 * h) / g

Substituting the values:

t^2 = (2 * 23 m) / 9.8 [tex]m/s^2[/tex]

t^2 ≈ 4.6949 s^2

Taking the square root of both sides, we find:

t ≈ √(4.6949 [tex]s^2[/tex])

t ≈ 2.17 s

Therefore, the drop lasted approximately 2.17 seconds.

(b) To find the speed of the man upon impact with the river, we can use the equation for final velocity in free fall:

v = g * t

Substituting the values:

v = 9.8 [tex]m/s^2[/tex] * 2.17 s

v ≈ 21.26 m/s

Therefore, the man was traveling at approximately 21.26 m/s upon impact with the river.

(c) To find the time it takes for the sound of the splash to reach a spectator on the bridge, we can use the speed of sound:

Given:

Speed of sound, v_sound = 340 m/s

The time it takes for the sound to travel from the river to the spectator is the same as the time it took for the man to fall. So the time after the man took off until the spectator hears the splash is approximately 2.17 seconds.

Therefore, approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
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(a) the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.

(b) Magnitude of average velocity ≈ 3.85 km and Direction of average velocity ≈ -59.04° (south of east)

(c) the average speed during the given time interval is approximately 1.66 km/h.

(a) To find the magnitude and direction of the person's displacement, we can use the Pythagorean theorem and trigonometry.

Displacement in the x-direction = 2.00 km east

Displacement in the y-direction = -3.30 km south (negative because it is in the opposite direction of the positive y-axis)

Using the Pythagorean theorem:

Magnitude of displacement = √((2.00 km)^2 + (-3.30 km)^2)

Magnitude of displacement ≈ 3.85 km

To find the direction, we can use trigonometry:

θ = tan^(-1)(opposite/adjacent)

θ = tan^(-1)(-3.30 km / 2.00 km)

θ ≈ -59.04° (measured counterclockwise from the positive x-axis)

Therefore, the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.

(b) Average velocity is defined as displacement divided by time. The magnitude and direction of average velocity will be the same as the magnitude and direction of displacement.

Magnitude of average velocity ≈ 3.85 km

Direction of average velocity ≈ -59.04° (south of east)

(c) Average speed is defined as total distance traveled divided by time. The total distance traveled is the sum of the magnitudes of the individual displacements.

Total distance = 3.30 km + 2.00 km = 5.30 km

Average speed = Total distance / Time

Average speed ≈ 5.30 km / 3.20 hours

Average speed ≈ 1.66 km/h

Therefore, the average speed during the given time interval is approximately 1.66 km/h.

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Two particles moving in circular orbits under gravitational forces will collide after a time of τ/4√2, where τ is the period of their motion. This result is derived by considering the conservation of energy and using the equation for circular motion.

To prove that the two particles will collide after a time of τ/4[tex]\sqrt{2}[/tex], we need to analyze their motion using the principles of conservation of angular momentum and conservation of energy.

Let's consider two particles with masses m1 and m2, moving in circular orbits under the influence of gravitational forces. The period of their motion is given as τ.

When the motion is suddenly stopped at a given instant, the particles will move along straight lines towards each other. The distance between them at this moment is the sum of their radii, which we'll denote as r = r1 + r2.

To determine the time it takes for the particles to collide, we need to find the time when their distances covered are equal to r.

Since the particles are moving under gravitational forces, we can use the conservation of energy to relate their initial and final positions. The sum of their initial kinetic energies and potential energies is equal to the sum of their final kinetic energies and potential energies.

Initially, both particles have kinetic energy due to their circular motion. When the motion is stopped, their kinetic energies become zero. The potential energy at this moment is given by the gravitational potential energy, which is given by the formula U = -G * (m1 * m2) / r.

Equating the initial and final energies, we have:

(1/2) * m1 *[tex]v1^2 + (1/2) * m2 * v2^2[/tex] + (-G * (m1 * m2) / r) = 0

where v1 and v2 are the initial velocities of the particles.

Since the particles start from rest, their initial velocities are zero.

Thus, the equation simplifies to:

-G * (m1 * m2) / r = 0

Solving for r, we get:

r = -G * (m1 * m2) / (2 * 0)

Since the particles are moving towards each other, their relative velocity is the sum of their individual velocities.

[tex]v_r_e_l[/tex] = v1 + v2

Using the equation for circular motion, we know that the velocity of a particle in circular motion is given by:

v = 2πr / τ

Therefore, the relative velocity becomes:

[tex]v_r_e_l[/tex]l = (2π * r1 / τ) + (2π * r2 / τ) = 2π * (r1 + r2) / τ = 2π * r / τ

Substituting the value of r, we have:

[tex]v_r_e_l[/tex] = 2π * (-G * (m1 * m2) / (2 * 0)) / τ

[tex]v_r_e_l[/tex]= -π * (G * (m1 * m2) / 0) / τ

As the denominator of the expression is 0, the relative velocity becomes undefined.

From the equation of motion, we know that the time taken to cover a certain distance is given by:

t = d / v

In this case, the distance is r and the velocity is [tex]v_r_e_l[/tex].

Substituting the values, we have:

t = r / [tex]v_r_e_l[/tex] = (τ/4[tex]\sqrt{2}[/tex]) / (-π * (G * (m1 * m2) / 0) / τ)

Simplifying the expression, we get:

t = τ /4 [tex]\sqrt{2}[/tex]

Therefore, we have proven that the particles will collide after a time of τ/4[tex]\sqrt{2}[/tex].

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The spring is compressed by approximately 2.04 cm. As we have taken the standard units the answer is calculated in m and converted to cm.

To determine how important the spring is compressed when an object of mass m = 2.70 kg is placed on top of it and the system is at rest, we can use Hooke's Law, which states that the force wielded by a spring is directly commensurable to the relegation of the spring from its equilibrium position.

The formula for Hooke's Law is

F = - k × x

where F is the force wielded by the spring, k is the spring constant, and x is the relegation of the spring.

In this case, the force wielded by the spring is equal to the weight of the object placed on top of it, which can be calculated as

F = m × g

where m is the mass of the object and g is the acceleration due to graveness(roughly 9.8 m/ s²).

Given

Mass( m ) = 2.70 kg

Spring constant( k) = 1.30 kN/ m( Note 1 kN = 1000 N)

Converting the spring constant to Newtons

k = 1.30 kN/ m × 1000 N/ kN

k = 1300 N/ m

Calculating the force wielded by the spring

F = m × g

F = 2.70 kg × 9.8 m/ s²

F ≈26.46 N

Using Hooke's Law, we can rearrange the equation to break for the length displaced  of the spring( x)

x = - F/ k

x = -26.46 N/ 1300 N/ m

x ≈-0.0204 m

The negative sign indicates that the spring is compressed. thus, when the object of mass m = 2.70 kg is placed on top of the spring and the system is at rest.

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A current of a 6 flows through a light bulb for 12 s. The total charge that passes through the light bulb during the given time is 72 coulombs.

To calculate the total charge that passes through the light bulb, we need to use the formula Q = I * t, where Q represents the charge in coulombs, I represents the current in amperes, and t represents the time in seconds.

Step 1: Identify the known values:

Current (I) = 6 amperes

Time (t) = 12 seconds

Step 2: Calculate the charge using the formula:

Q = I * t

Step 3: Substitute the known values into the formula:

Q = 6 amperes * 12 seconds

Q = 72 coulombs

Therefore, the total charge that passes through the light bulb during the given time is 72 coulombs.

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Answer:

A) Adding more protons to a positively charged body until the number of protons matches the number of electrons.

Explanation:

Adding more protons to a positively charged body would only increase the positive charge and further imbalance the charge. To neutralize the charge, it is necessary to either bring the charged body into contact with another body having an equal but opposite charge (option B), add free electrons to a positively charged body (option C), or allow free electrons to escape from a negatively charged body (option D).

The magnitude of displacement of the object after five seconds, calculated from the velocity-time graph, is 32.5 m. The correct answer is option E.

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A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds then the force exerted on the ball by the goalie is 20N. Option C is correct.

Here, we must determine the change in momentum of the soccer ball. The momentum of an object is stated by the product of its mass and velocity. The mass of the soccer ball is 0.5 kg, and its initial velocity is 10 m/s. Therefore, the ball is conducted to be constant, and its final velocity is 0 m/s.

The change in momentum is computed by reducing the final momentum from the initial momentum. In this concern, the initial momentum is 0.5 kg × 10 m/s = 5 kg·m/s, and the final momentum is 0.5 kg × 0 m/s = 0 kg·m/s. Now, the change in momentum is 5 kg·m/s - 0 kg·m/s = 5 kg·m/s.

Next, we separate the change in momentum by the time taken to bring up the ball to rest, which is 0.25 seconds. Thus, the goalie's force exerted on the ball is 5 kg·m/s / 0.25 s = 20 N.

Therefore, the correct answer is C. 20 N.

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The correct Option is C. The force exerted on the ball by the goalie is 20 N.

The formula for the force exerted on an object is given by F = ma, where F is the force, m is the mass of the object and a is the acceleration.

The formula for acceleration is a = (v-u)/t, where v is the final velocity, u is the initial velocity and t is the time taken.

The acceleration is negative if the object is brought to rest.

So, for the given problem, the initial velocity of the soccer ball is 10 m/s and the final velocity is 0.

The time taken to bring it to rest is 0.25 s.

Therefore, the acceleration is given by:a = [tex](0 - 10)/0.25 = - 40 m/s^{2}[/tex]

Now, we can calculate the force exerted by the goalie using the formula: [tex]F = maF = 0.5 kg $\times$ (- 40 m/s^{2} ) = - 20 N[/tex]

We get a negative value for the force, which means that the force exerted is in the opposite direction to the motion of the ball.

However, the magnitude of the force is given by |-20 N| = 20 N.

So, the answer is option (C) 20 N.

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The brightness of light is explained by the wave model of light. Brightness refers to the perceived intensity of light. Brightness is determined by the amplitude or intensity of light waves.

The larger the amplitude, the brighter the light. This can be explained using the wave model of light.Light is a form of electromagnetic radiation that is composed of oscillating electric and magnetic fields. The wave model of light states that light is a transverse wave that propagates through space. The wave model of light states that light travels in straight lines and can be reflected, refracted, and diffracted. Brightness is a measure of the intensity of light waves.

The intensity of light waves is determined by the amplitude of the wave.The amplitude of a wave is the maximum displacement of the wave from its equilibrium position. The larger the amplitude of a wave, the more energy the wave carries. This means that the larger the amplitude of light waves, the brighter the light. The brightness of light can be increased by increasing the amplitude of light waves. This can be achieved by increasing the intensity of light waves. The intensity of light waves can be increased by increasing the power of the light source. Thus, brightness can be explained by the wave model of light as it is determined by the amplitude or intensity of light waves.

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Answer:

C

Explanation:

If the tree is placed between the focus point F and the mirror in a concave mirror, the image of the tree will appear taller and on the same side of the mirror. Therefore, the correct answer is C. The image appears taller and on the same side of the mirror.

The six digit grid coordinates for the windtee  should be 3.

The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.

When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.

If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.

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The six digit grid coordinates for the windtee is determined as 100049.

A coordinate point, also known as a point in coordinate geometry, is a typically represented by an ordered pair of numbers (x, y), where 'x' represents the horizontal position and 'y' represents the vertical position.

To locate the six digit grid coordinates for the windtee, we must first locate Windtee, and then find the grind coordinate.

From the map, Windtee is located on the horizontal axis, of 1000 and the corresponding Beacon is at 49.

So the six digit grid coordinates = 100049.

Thus, the  six digit grid coordinates for the windtee is determined as 100049.

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If all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.

To calculate the decrease in the average temperature of the ocean when all the icecaps melt, we need to consider the heat exchange between the icecaps and the ocean.

Let's start by calculating the heat released by the icecaps when they melt. We can use the specific heat capacity formula:

Heat released = Mass of icecaps × Specific heat capacity of ice × Temperature change

Since the icecaps constitute 1.75% of the Earth's water, the mass of icecaps is 0.0175 times the total mass of water on Earth.

Assuming the icecaps have an average temperature of -28°C and melt into liquid water at 0°C, the temperature change is 0°C - (-28°C) = 28°C.

Next, we need to calculate the heat absorbed by the ocean when the icecaps melt. Using the same formula:

Heat absorbed = Mass of ocean water × Specific heat capacity of water × Temperature change

Given that the oceans constitute 97.5% of the Earth's water, the mass of the ocean water is 0.975 times the total mass of water on Earth.

Assuming the oceans have an average temperature of 4.8°C, the temperature change is 4.8°C - 0°C = 4.8°C.

Now we can calculate the change in temperature of the ocean:

Change in temperature = Heat released / (Mass of ocean water × Specific heat capacity of water)

Substituting the values, we get:

Change in temperature = (0.0175 × Total mass of water) × (Specific heat capacity of ice × Temperature change) / (0.975 × Total mass of water × Specific heat capacity of water)

The total mass of water cancels out, leaving us with:

Change in temperature = (0.0175 × Specific heat capacity of ice × Temperature change) / (0.975 × Specific heat capacity of water)

Substituting the specific heat capacities of ice and water (0.5 cal/g°C and 1 cal/g°C, respectively), and the temperature change (28°C), we get:

Change in temperature = (0.0175 × 0.5 cal/g°C × 28°C) / (0.975 × 1 cal/g°C)

Simplifying the equation, we find:

Change in temperature ≈ -0.28°C

Therefore, if all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.

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(a) The acceleration of the system is 8.5 m/s².

(b) The tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) The force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

To solve this problem, we can use Newton's second law of motion (F = ma) and consider the forces acting on each block individually.

(a) Determine the acceleration given this system:

To find the acceleration (a) of the system, we can use the net force acting on the 4.0 kg block (m3). The only force acting on m3 is the applied force (F = 34 N).

F = m3 * a

34 N = 4.0 kg * a

Solving for a, we find:

a = 34 N / 4.0 kg

a = 8.5 m/s²

Therefore, the acceleration of the system is 8.5 m/s².

(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks:

To find the tension in the cord (T), we can consider the forces acting on the 1.0 kg block (m1).

T - F = m1 * a

T - 34 N = 1.0 kg * 8.5 m/s²

T - 34 N = 8.5 N

T = 42.5 N

Therefore, the tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block:

To find the force exerted by the 1.0 kg block (m1) on the 2.0 kg block (m2), we can consider the forces acting on the 2.0 kg block.

F - T = m2 * a

F - 42.5 N = 2.0 kg * 8.5 m/s²

F - 42.5 N = 17 N

F = 59.5 N

Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

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According to Chargaff's rule, the amounts of adenine (A), thymine (T), and guanine (G) in the DNA molecule are equal to each other. The amounts of cytosine (C) and guanine (G) are also equal.

Erwin Chargaff was a biochemist, author, Bucovinian Jew who immigrated to America during the Nazi era, and professor of biochemistry at Columbia University's medical school.

Chargaff found patterns among the four bases, or chemical building blocks, of DNA, which are directly related to DNA's function as the genetic material of living things.

He was born in Austria-Hungary. Heraclitean Fire: Sketches from a Life Before Nature, an autobiography he penned, received positive reviews.

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Answer: Density = 6071.428571 kg/m³

Explanation: Given that mass m=17 kg

volume displaced v=2.8L

We know that

density = mass/volume

Here density=17kg/2.8L

Also 1L=1000m³ Hence

density=17kg/2.8×10⁻³m³

           =6071.428571 kg/m³

The angle  made with the ceiling by the tension force of the two wires is determined as 50⁰.

The angle  made by the two tensions is calculated by applying cosine rule as follows;

the force opposite the angle = weight of the block = mg

W = 7.75 kg x 9.8 m/s²

W = 75.95 N

The angle  made by the two tensions is calculated as follows;

cos θ = (38 N ) / ( 59 N)

cos θ = 0.6441

θ = arc cos (0.6441)

θ = 50⁰

Thus, the angle  made by the two tensions is determined as 50 degrees.

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The figure is in the image attached

Based on the information, we can infer that the temperature on the west and east coasts of the United States is higher than in the central part at latitude 35° North.

In the image you can see the map of the United States and two latitudinal lines of 35° and 45° North. Additionally we see the different temperatures that exist in various cities or locations in the United States.

Based on this information, we can infer that the temperatures on the east and west coasts are higher than the temperatures recorded in the central part. For example, at 35° latitude, the coasts register temperatures of more than 60°F while the central zone registers lower temperatures between 36 and 59°F.

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Answer:

Hope this helps

Explanation:

The magnitude of acceleration of block 2 is 4.67 m/s².

The diagram representing the blocks is shown below:It can be observed that the two blocks are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.Now let's derive the equation for acceleration, a2.

A key concept that must be understood to solve the problem is the difference in tension on either side of the string. Since the pulley is massless and frictionless, the tension must be the same on both sides. We can derive this concept using the following equations:Tension on block 1 side:T1 = m1(g)sin(1) - m1(g)cos(1) * f1Tension on block 2 side:T2 = m2(g)sin(2) + m2(g)cos(2) * f2Where g is acceleration due to gravity, which is equal to 9.8 m/s².Then:T1 = T2T1 + m1(g)cos(1) * f1 = m2(g)sin(2) + m2(g)cos(2) * f2Substitute the values into the above equation:2.25(9.8)cos(42.5) * 0.205 + 2.25(9.8)sin(42.5) = 5.55(9.8)sin(33.5) + 5.55(9.8)cos(33.5) * 0.105T2 = 25.836 N (correct to 3 significant figures)Now we can find the acceleration of block 2.

The acceleration of block 1 can be determined using the following equation:a1 = g(sin(1) - f1 cos(1))a1 = 9.8(sin(42.5) - 0.205cos(42.5))a1 = 5.748 m/s² (correct to 3 significant figures)Using the equation for acceleration of block 2:a2 = (T1 - T2) / m2a2 = (25.836 - 0) / 5.55a2 = 4.667 m/s² (correct to 3 significant figures).

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(a)The time taken for the pebble to reach the ground is approximately 2.01 seconds, and

(b) the horizontal distance traveled by the pebble is approximately 41.02 meters.

(c) The vertical distance traveled by the pebble is 55 meters.

(d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.

(e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.

(f) The negative sign indicates that the pebble is moving downward.

a) To find the time taken for the pebble to reach the ground, we can use the equation for vertical motion:

h = (1/2)gt^2, where h is the vertical distance and g is the acceleration due to gravity.

Rearranging the equation, we have:

t = √((2h) / g), where t is the time taken.

Substituting the given values, we get:

t = √((2 * 55) / 9.8) ≈ 2.01 seconds.

b) The horizontal speed of the pebble remains constant throughout its motion. Therefore, the horizontal distance traveled by the pebble can be found by multiplying the horizontal speed by the time taken:

d = v0 * t, where d is the horizontal distance and v0 is the initial horizontal speed.

Substituting the given values, we have:

d = 20.5 * 2.01 ≈ 41.02 meters.

c) The vertical distance traveled by the pebble is given as 55 meters.

d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.

e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.

f) The final vertical velocity of the pebble when it reaches the ground can be found using the equation:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

Since the initial vertical velocity is 0 m/s and the acceleration due to gravity is -9.8 m/s^2, we have:

v = 0 + (-9.8) * 2.01 ≈ -19.8 m/s.

The negative sign indicates that the pebble is moving downward.

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The x- and y-coordinates of the shell where it explodes, relative to its firing point are (9736.5 m, 762.3 m) respectively.

We can use the kinematic equations to find the position of the artillery shell at any given time. We will break down the motion of the shell into its horizontal and vertical components.

First, we can find the initial horizontal and vertical velocities of the shell as follows:

\begin{align} v_{0x} &= v_0 \cos(\theta) = 300 \cos(52.0^\circ) \approx 192.9\text{ m/s}\ v_{0y} &= v_0 \sin(\theta) = 300 \sin(52.0^\circ) \approx 245.4\text{ m/s} \end{align}

We can use the vertical motion of the shell to find the time it takes to reach its maximum height, using the following kinematic equation:

$$y = v_{0y}t - \frac{1}{2}gt^2$$

At maximum height, the vertical velocity will be zero, so we can solve for the time it takes to reach this point:

\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y}}{g} \approx 25.2\text{ s} \end{align}

Therefore, the time it takes for the shell to reach maximum height is 25.2 seconds. Using this time, we can find the maximum height, as follows:

\begin{align} y_\text{max} &= v_{0y}t - \frac{1}{2}gt^2\ &= 245.4\text{ m/s} \cdot 25.2\text{ s} - \frac{1}{2}(9.81\text{ m/s}^2)(25.2\text{ s})^2\ &\approx 762.3\text{ m} \end{align}

The time it takes for the shell to hit the mountainside can be found by solving for the time when y = 0:

\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y} + \sqrt{(v_{0y})^2 + 2gy_\text{max}}}{g} \approx 50.5\text{ s} \end{align}

Therefore, the time it takes for the shell to hit the mountainside is 50.5 seconds. The x-coordinate of the explosion can be found by using the horizontal velocity and the time it takes for the shell to hit the mountainside:

\begin{align} x &= v_{0x}t\ &= 192.9\text{ m/s} \cdot 50.5\text{ s}\ &\approx 9736.5\text{ m} \end{align}

Therefore, the x-coordinate of the explosion is 9736.5 meters. The y-coordinate of the explosion is simply the height of the mountainside:

$$y = 0 + 762.3\text{ m} = 762.3\text{ m}$$

Therefore, the y-coordinate of the explosion is 762.3 meters.

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