What can develop when the temperature of the ocean is 80°F. A. Tornado B. Tsunami C. Thunder Storm D. Hurricane

If you were standing still on the equator of Earth you would move at nearly 1,000 miles/hour which is approximately 1,600 km/hr.

What is speed?

Speed is measured as the ratio of distance to the time in which the distance was covered.

The Earth is round and spins once every 24 hours. This spinning of the earth causes the daily motion which we feel here. If you were standing still on the equator of Earth you would move at nearly 1,600 km/hr.

Speed is a scalar quantity as it has only direction and no magnitude.

Speed is defined mathematically as distance/ time.

speed = distance/time expressed in m/s.

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The space cruiser is 11070000 m far from the planet's surface

The following data were obtained from the question:

We can obtain the distance of the space cruiser from the planet's surface as follow:

s = ut + ½gt²

s = (0 × 600) + (½ × 61.5 × 600²)

s = 0 + (½ × 61.5 × 360000)

s = 0 + 11070000

s = 11070000 m

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The electric force on one of the masses is 0.6 N.

The acceleration of the mass is 0.35 m/s².

The electric force between the masses is calculated as follows;

F = kq²/r²

where;

F = (9 x 10⁹ x (9.8 x 10⁻⁶)²)/(1.2²)

F = 0.6 N

The acceleration of the mass is calculated as follows;

F = ma

a = F/m

a = (0.6 N) / (1.7 kg)

a = 0.35 m/s²

Thus, the electric force on one of the masses is 0.6 N.

The acceleration of the mass is 0.35 m/s².

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The distance travelled by Ann before she begins to slow down is 11.65 m

First, we shall enlighten ourselves on what speed is. This is given below.

Speed is the distance an object travelled per unit time. It can be expressed as:

Speed = distance / time

Finally, we can obtain the distance Ann travelled as illustrated below.

From the question given above, the following data were obtained:

Speed = distance / time

15.56 = distance / 0.749

Cross multiply

Distance = 15.56 × 0.749

Distance = 11.65 m

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The direction of the initial acceleration vector will point towards the wall, and its magnitude will be less than the acceleration vector of the crash, therefore the correct answer is option A.

The rate of change of the velocity with respect to time is known as the acceleration of the object.

As given in the problem, a test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction of the initial acceleration vector and compare the initial acceleration vector's magnitude with respect to the crash acceleration magnitude.

Thus, the initial acceleration vector will point in the direction of the wall and be smaller than the crash's acceleration vector, therefore the correct answer is option A.

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The car travels 25 meters before coming to rest and will hit the deer.

Given in the Question,

Initial speed = u = 15 m/s

Deceleration = a = 4.5 m/s²

According to the question, if the car stops before traveling 20m, it will avoid hitting the deer. So, we need to find the stopping distance for the car.

Deceleration is negative acceleration, so the sign of acceleration is will be -.

Also, the car comes to rest after applying the brakes. Therefore the final speed of the car will be zero.

Final speed = v = 0 m/s

Using the Third equation of motion,

v² - u² = 2as

Put in the values, we get

(0)² - (15)² = 2(-4.5)s

-2×4.5 × s = - 15× 15

s = 225/9

s = 25 m

So the car will come to rest after traveling 25 meters. But the Deer is present at 20 meters; therefore the car will hit the deer.

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The correct answer is :

Here 100 N force is applied to make the box move with constant velocity from rest. That means 100 N force is applied to overcome the limiting  static friction and as soon as 100 N force is applied it starts moving.

Now,

Constant velocity means acceleration = 0

Net force acting on the box =mass × accelaration = mass × 0 = 0

Conceptually it is zero as it is balanced by kinetic friction which has equal value that of applied force. Because net force =Applied force - friction force and hence here friction force =applied force.

If there was any accelaration then there would exist a net force and then frictional force and applied force will be the same.  

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Alexus was resting on the couch but then decided she needed some food. She ran at up to 15m/s, accelerating at 3m/s² the whole time, until she got to the fridge . The fridge will be 37.5 m far away from the couch

The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Equation of kinematics  are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.

given

final velocity = 15m/s

initial velocity = 0

a = 3m/s²

s = ?

a = acceleration = final - initial / time

3 = 15 - 0 / time

time = 15 /3 = 5 seconds

[tex]v^{2}[/tex] - [tex]u^{2}[/tex] = 2as

[tex]15^{2}[/tex] - 0 = 2 * 3 * s

s = 225 /6

  = 37.5 m

The fridge will be 37.5 m away from the couch

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Answer:

58.8 metres

Explanation:

using the formula

h=ut+½gt

U=0

g=9.8m/s²

t=12.0a

Hence, h= 0.5 × 9.8 × 144

h = 58.8 metres

Explanation:

probably luck or the right material's

Answer: I believe that the answer to your question is "Inertia", or A

Explanation:

The height of the 2nd spot is 3 times the height of the 1st spot (Option A)

We'll begin by obtaining the height of the first spot. This can be obtained as follow:

v = 2x / t

343 = 2x / 4

Cross multiply

2x = 343 × 4

2x = 1372

Divide by 2

x = 1372 / 2

Height of first spot = 686 m

v = 2x / t

343 = 2x / 12

Cross multiply

2x = 343 × 12

2x = 4116

Divide by 2

x = 4116 / 2

Height of second spot = 2058 m

Comparison = Height of 2nd spot / Height of 1st spot

Height of 2nd spot / Height of 1st spot = 2058 / 686

Height of 2nd spot / Height of 1st spot = 3

Cross multiply

Height of 2nd spot = 3 × Height of 1st spot

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Answer: Vx is changing  and  Vx is greatest near the beginning of the launch.

Explanation: Think of Vx as a bullet fired from a gun. As soon as it is fired it starts to slow down so Vx is changing and at the beginning Vx has its greatest velocity because it can only go slower from there.

answers are B and C. horizontal velocity never changes so A cannot be the answer. that makes B one of the answers. I would also say that the horizontal velocity is greatest near the beginning of the launch as it just starts, since it stays at a constant pace after, so D is not an answer. hope this helps ^_^

The magnitude of the change in velocity is determined as 408.94 km/h.

The change in direction of the velocity is 64.4⁰ north of west.

The magnitude of the average acceleration during the trip is 0.0105 m/s².

The magnitude of change in velocity is the resultant velocity of the plane.

v² = a² + b² - 2ab cosθ

where;

v² = (192²) + (250²) - 2(192 x 250) cos(135)

v² = 167,236

v = √167,236

v = 408.94 km/h

vyi = 192 km/h

vy2 = 250 x sin(45) = 176.77 km/h

vy(total) = 192 km/h + 176.77 km/h = 368.77 km/h

vxi = 0

vx2 = - 250 km/h x cos(45) = -176.77 km/h

θ = arc tan (Vy/Vx)

θ = arc tan(368.77 / -176.77)

θ = -64.4 ⁰

θ = 64.4⁰ north of west.

a = v/t

v = 408.94 km/h = 113.6 m/s

h = 3 h = 10,800 seconds

a = (113.6 m/s) / ( 10,800 s)

a = 0.0105 m/s²

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Answer:

A)  4.95 m

B)  1.48225 m

Explanation:

Constant Acceleration Equations (SUVAT)

[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]  

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Consider the horizontal and vertical motion of the projectile separately.

The horizontal component of velocity is constant, as there is no acceleration horizontally.

Resolving horizontally, taking → as positive:

[tex]u=9.00\quad v=9.00 \quad a=0\quad t=0.550[/tex]

[tex]\begin{aligned}\textsf{Using} \quad s & = \left(\dfrac{u+v}{2}\right)t:\\\\s&= \left(\dfrac{9+9}{2}\right)(0.550)\\s&= (9)(0.550)\\ \implies s&= 4.95\:\sf m\\\end{aligned}[/tex]

As the projectile is fired horizontally, the vertical component of its initial velocity is zero.

Acceleration due to gravity = 9.8 ms⁻²

Resolving vertically, taking ↓ as positive:

[tex]u=0\quad a=9.8\quad t=0.550[/tex]

[tex]\begin{aligned}\textsf{Using} \quad s & = ut+\dfrac{1}{2}at^2:\\\\s&= (0)(0.550)+\dfrac{1}{2}(9.8)(0.550)^2\\s&= 0+(4.9)(0.3025)\\\implies s&= 1.48225\:\sf m\\\end{aligned}[/tex]

The distance to the destination is 18,000 meters or 18 kilometers. the formula for calculating the distance is speed × Time.

Using the formula:

Distance = Speed × Time

Speed = 15 m/s (meters per second)

Time = 20 minutes

First, we need to convert the time from minutes to seconds, as the speed is given in meters per second.

1 minute = 60 seconds

Time in seconds = 20 minutes × 60 seconds/minute = 1200 seconds

Now, we can calculate the distance:

Distance = 15 m/s × 1200 seconds

Distance = 18,000 meters

So, the distance to the destination is 18,000 meters, or 18 kilometers (since 1 kilometer = 1000 meters).

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Answer: 1 > 3 > 2

Explanation: The range will increase with the velocity. If they are all launched at the same time the ones that are launched the hardest or with the most velocity will go the furthest horizontally. But because they are all launched from the same height and the force of gravity on all three projectiles is constant (the same for all three) they will all hit the ground at the same time but different distances from the starting point.

Answer:

Explanation:

The correct option will be the option (d) that is 1.09 A.

According to the equation given in the question:

That is; P=i^2R                  where I= current, R= resistance

           P=30 W

           R=25 ohm

            30=i^2*25

              I^2=30/25

             I^2=1.2

             I=1.09A

So, the correct answer of the given question will be 1.09A

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The frequency of a radiation that has a wavelength of 9.17 × 10-⁷m is 3.27 × 10¹⁴ Hz.

The frequency of a radiation can be calculated by dividing the velocity of the radiation by the wavelength. That is;

f = v/(λ)

According to this question, some radiation detected by a thermal imaging camera during a training exercise has a wavelength of 9.17 × 10-⁷m. The frequency can be calculated as follows:

f = 3 × 10⁸m/s ÷ 9.17 × 10-⁷m

f = 3.27 × 10¹⁴ Hz

Therefore, the frequency of a radiation that has a wavelength of 9.17 × 10-⁷m is 3.27 × 10¹⁴ Hz.

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The time of flight of the ball is 17.12 seconds.

The horizontal distance or range of the ball is 354.8 m.

The time of flight of the ball is calculated as follows;

T = (2u sinθ)/g

where;

T = (2 x 25 sin34) / (¹/₆ x 9.8)

T = 17.12 s

The range of the golf ball is calculated as follows;

R = Uxt

R = (U cosθ)t

R = (25 cos34) x 17.12

R = 354.8 m

Thus, the time of flight of the ball is 17.12 seconds.

The horizontal distance or range of the ball is 354.8 m.

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We consider an elementary mass located at this point to be in a stable equilibrium.

An equilibrium is said to be stable if small, externally induced displacements from that state produce forces that tend to oppose the displacement and return the body or particle to the equilibrium state.

Also a body is said to be in a stable equilibrium, if the opposing forces acting on a body are equal.

Thus, along the line through the centers of the Moon and the Earth, there is a point where the gravitational forces exerted by the Moon and the Earth, respectively, are equal. We consider an elementary mass located at this point to be in a stable equilibrium.

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You throw a rock from the upper edge of a 75.0 -m vertical dam with a speed of 25.0 m/s at 65.0∘ above the horizon.

The horizon is the line that, when seen from a position on or near the surface of a celestial body, appears to separate the surface from the sky of that body. All viewing directions are split according to whether it crosses the surface of the relevant body or not.

Since the true horizon is an imaginary line, it can only be seen with any degree of accuracy when it is situated along a generally flat surface, such as the oceans of the Earth. On Earth, the geography may also cause biological objects like trees and/or man-made objects like buildings to obstruct this line in some locations. The location where these obstructions overlap the sky is known as the visible horizon.

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The heigth of the cliff is 63.504 m.

Height is the vertical distance  between two points.

To calculate the height of the cliff, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the heigth of the cliff is 63.504 m.

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It will take the fugitive 4.83 s to catch the empty box car

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

The time taken for the fugitive to catch the car if he maintains his maximum speed can be obtained as follow:

a = (v – u) / t

Thus,

t = (v – u) / a

t = (5.8 – 0) / 1.2

t = 5.8 / 1.2

t = 4.83 s

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Answer:

A battery

Explanation:

A battery is the biggest conductor. I think.

Answer:

100

Explanation:

3-2=1

100/1=100