Answer:
Explanation:
charge on each proton = 1.6 x 10⁻¹⁹ C
mass of proton = 1.67 x 10⁻²⁷ kg
Electrostatic force of repulsion Fe = 9 x 10⁹ x ( 1.6 x 10⁻¹⁹ )² / d²
= 23.04 x 10⁻²⁹ / d²
Gravitational force of attraction = G Mm / d²
M = m = 1.67 x 10⁻²⁷ kg
Gravitational force of attraction Fg = 6.67 x 10⁻¹¹ x ( 1.67 x 10⁻²⁷ )² / d²
= 18.60 x 10⁻⁻⁶⁵ / d²
So Fg is far less than Fe and former is attractive , later is repulsive .
Fe > Fg, opposite direction , is the answer .
A bullet traveling at 5.0 x10^2 meters per is brought to rest by an impulse of 50 Newton*seconds. Find the mass of the bullet.
The bullet stops moving on hitting on a surface. Hence, the impulse here is equal to the momentum. Therefore, the mass of the bullet is 0.1 Kg.
What is impulse?Impulse in physics is the change in momentum. It is the product of the force and change in time.
hence, impulse = f. dt
When the bullet is travelling with a velocity of 500 m/s it has a momentum. When it brought to rest, momentum become zero. Thus, the momentum is equal to the impulse here.
Therefore, f.dt = m. v
f.dt = 50 N s
v = 500 m/s
m = 50 N s/500 m/s = 0.1 Kg
Therefore, the mass of the bullet is 0.1 Kg.
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Why is Energy, Work and Power all Scalar Quantity?
Answer:
Explanation:
We already know that Force is a vector. Weight being a force, is also a vector quantity. Displacement is distance in a specific direction, hence it is a vector quantity too. ... since energy, work done and time are scalar, Power is a scalar quantity
A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect friction between car and road. Answer in units of m/s.
Answer:
1.66 m/s
Explanation:
Work or kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex]
[tex]4864=\frac{1}{2} (3540)v^{2}[/tex]
v = 1.66 m/s
A supercluster is 100 million light-years across. How long would it take light to travel from one edge of the supercluster to the center of the supercluster?
Answer:
50 million years
Explanation:
light years is the distance light travels in one year given that the supercluster is 100 million light years across the the distance to the center will be half that amount therefore the answer is 50 million years
Kind of energy a piece of radioactive metal contains
Answer:
Radioactive materials give off a form of energy called ionizing radiation.
A iguana runs back and forth along the ground. The horizontal position of the iguana in meters over time is shown
What is the displacement of the iguana between 3 s and 6 s?
m
What is the distance traveled by the iguana between 3 s and 6 s?
The displacement of the iguana between 3 s and 6 s is 6.71 meters.
The distance traveled by the iguana between 3 s and 6 s is 8.08 meters.
What are distance and displacement?Distance is the sum of an object's movements, regardless of direction.
The term "displacement" refers to a shift in an object's position.
According to the graph:
The displacement of the iguana between 3 s and 6 s
= √{ (3-6)²+(6-0)²} meters
= 6.71 meters.
The distance traveled by the iguana between 3 s and 6 s
= [ √{ (3-5)²+(6-6)²} +√{ (5-6)²+(6-0)²}] meters
= [2+ 6.08] meters
= 8.08 meters.
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A 2kg block is attached to a spring for which K=200N/m it is held at an extension of 5 cm and then released at t=0.
A, the displacement as a function of time?
B, the acceleration when X=+A/2
C, the total energy when X=+A/2
D, the velocity when X=+A/2
The displacement, acceleration, energy and velocity of the simple harmonic motion of the mass attached to the spring are as follows;
A) x(t) = 0.05·sin(10·t + π/2)
B) The acceleration is; a(t) = -2.5 m/s²
C) The total energy is 0.0625 J
D) The velocity is ±√3/4 m/s
What is a simple harmonic motion?The restoring force of a body in simple harmonic motion is directly proportional to the displacement of the body from its mean or central position.
Mass of the block, m = 2 kg
The spring constant, k = 200 N/m
The extension of the spring = 5 cm
Time at which the spring is released, t = 0
A. The motion of the spring with the mass is a Simple Harmonic Motion
The angular velocity can be obtained using the formula;
ω = √(k/m)
Therefore;
ω = √(200/2) = 10
The angular velocity of the block on the spring is, ω = 10 rad/s
The period, T = The time to complete 2·π rad
Therefore; T = 2·π rad/(10 rad/s) = π/5 s
The amplitude, A, is the cistance of the mass from the at rest position, which is 5 cm = 0.05 m
The equation of the extension of the spring is therefore;
x(t) = 0.05·sin(10·t + c)
At t = 0, x(t) = 0.05, therefore;
sin(10 × 0 + c) = sin(c) = 1
c = π/2
The equation for the displacement as a function of time is therefore;
x(t) = 0.05·sin(10·t + π/2)B. The acceleration when x(t) = A/2 is obtained as follows;
x(t) = 0.05·sin(10·t + π/2)
A/2 = 0.05·sin(10·t + π/2)
A = 0.05
0.05/2 = 0.05·sin(10·t + π/2)
sin(10·t + π/2) = 1/2
10·t + π/2 = π/6
t = -π/30
cos(10×(-π/30) + π/2) = ±√3/2
v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)
a(t) = v'(t) = -5·sin(10·t + π/2)
a(t) = v'(t) = -5·sin(10·t + π/2) = -5 × 1/2 = -2.5
The acceleration when X = + A/2 is -2.5 m/s²C. The energy in a pring = (1/2)·k·x²
When x = A/2, we get;
E = (1/2) × 200 × (0.05/2)² = 0.0625
The energy in the spring when x = A/2 is 0.0625 JD) The velocity when x = A/2 is; v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)
v(t) = 0.5 × cos(10·t + π/2)
When x = A/2, sin(10·t + π/2) = 1/2, therefore;
cos(10·t + π/2) = ±√3/2
v(t) = 0.5 × ±√3/2 = ±√3/4
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3 ) find the electrical force between the two charges Q1=3mc ,Q2=-6mc when they are 0.3 m parted ?
find the amount of the force when Q1 is doubled ?
Answer:
F = 3.6 10⁶ N
Explanation:
The expression for the electric force is
F = [tex]k \ \frac{q_1 q_2}{r^2}[/tex]
in this case it indicates that the charge q1 is doubled
q₁ = 2 3 10⁻³ C
q₁ = 6 10⁻³ C
let's reduce the magnitudes to the SI system
q₂ = 6 10⁻³ C
r = 0.3 m
let's calculate
F = 9 10⁹ 6 10⁻³ 6 10⁻³ / 0.3²
F = 3.6 10⁶ N
It’s bungee jumping skydiving and hiking
A dog drags a 1-kg bone across the floor for 2 meters with an applied
force of 10N. How long did it take him if he used 40 Watts of power?
seconds
Answer:
0.5 seconds
Explanation:
Work = Fdcos(theta) = 10*2*cos(0) = 20 J
Power = W/t
40 = 20/t
t = 0.5s
A satellite of mass m is orbiting Earth in a stable circular orbit of radius R. The mass and radius of Earth are ME and RE , respectively. Express your answers to parts (a), (b), and (c) the following in terms of m, R, ME , RE , and physical constants, as appropriate.
a. Derive an expression for the speed of the satellite in its orbit.
b. Derive an expression for the total mechanical energy of the satellite-Earth system in its orbit.
c. Derive an expression for the period of the satellite’s orbit.
Answer:
a) v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex], b) Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex], c) T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]
Explanation:
a) For this exercise we must use Newton's second law with the gravitational force
F = ma
[tex]G \frac{m M_e}{(R+R_e)^2 }[/tex] = m a
the acceleration of the satellite is centripetal
a = [tex]\frac{v^2}{(R+R_e)}[/tex]
we substitute
[tex]G \frac{m M_e}{(R+R_e)^2 } = m \frac{v^2}{ (R+R_e)}[/tex]
[tex]G \frac{M_e}{(R+R_e)}[/tex] = v²
v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex]
the distance is from the center of the earth
b) mechanical energy is the sum of kinetic energy plus potential energy
Em = K + U
Em = ½ m v² - G m M / (R + R_e)
we substitute the expression for the velocity
Em = ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex] - [tex]G \frac{M_e}{(R+R_e)}[/tex]
Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex]
c) as the orbit is circulating, the velocity modulus is constant
v = d / t
in a complete orbit the distance traveled of the circle is
d = 2π (R + R_e)
where time is called period
v = 2π (R + R_e)
T = 2π (R + R_e) / v
we substitute the speed value
T = 2π (R + R_e) [tex]\sqrt{\frac{(R+R_e) }{G M_e } }[/tex]
T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]
(a) An expression for the speed of the satellite in its orbit.
[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]
(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.
[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]
(c) An expression for the period of the satellite’s orbit.
[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]
What are satellites?A satellite is a moon, planet or machine that orbits a planet or star. For example, Earth is a satellite because it orbits the sun
a) For this exercise we must use Newton's second law with the gravitational force
F = ma
[tex]ma =G\sqrt{\dfrac{mM_e}{(R+R_e)}[/tex]
the acceleration of the satellite is centripetal
[tex]a=\dfrac{v^2}{R+R_e}[/tex]
we substitute
[tex]G\dfrac{mM_e}{(R+R_e)}=m\dfrac{v^2}{(R+R_e)}[/tex]
[tex]G\dfrac{M_e}{(R+R_e)}=v^2[/tex]
[tex]v=\sqrt{G\dfrac{M_e}{(R+R_e)}[/tex]
b) mechanical energy is the sum of kinetic energy plus potential energy
Em = K + U
[tex]Em =\dfrac{1}{2}m v^2 - \dfrac{G m M} {(R + R_e)}[/tex]
we substitute the expression for the velocity
[tex]E_m=\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}-G\dfrac{M_e}{(R+R_e)}[/tex]
[tex]E_m=-\dfrac{1}{2}G\dfrac{M_e}{(R+R_e)}[/tex]
c) as the orbit is circulating, the velocity modulus is constant
[tex]v=\dfrac{d}{t}[/tex]
in a complete orbit the distance traveled of the circle is
[tex]d = 2\pi (R + R_e)[/tex]
where time is called period
[tex]v = 2\pi (R + R_e)[/tex]
[tex]T = \dfrac{2\pi (R + R_e)} { v}[/tex]
we substitute the speed value
[tex]T = 2\pi (R + R_e) . \sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]
[tex]T=2\pi\sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]
(a) An expression for the speed of the satellite in its orbit.
[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]
(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.
[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]
(c) An expression for the period of the satellite’s orbit.
[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]
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A 6 kg bowling ball is lifted 1.2 m into a storage rack. The acceleration of gravity is 9.8 m/s² Calculate the increase in the ball's potential energy. Answer in units of J.
Answer:
70.56 J
Explanation:
Gravitational Potential Energy= mass×gravitational pull× height
= 6×9.8×1.2= 70.56 J
HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!
Answer:
a. 6000J or 6KJ
b. Force =600
Compare scalar and vector quantities using the definitions of distance and displacement
Answer:
Distance is a scalar quantity while displacement is a vector quantity
Explanation:
A scalar quantity represents only the magnitude and does not give any detail about the direction of the quantity for example distance. Distance can be any length measured in any direction (no specific direction)
However, a vector quality represents both the magnitude and direction. For instance displacement is a vector quantity. If direction is not defined then displacement becomes equal to distance.
in v-belts the contact between the pulley and the belt is at the
Answer:
Is at the pivot of the wheel
Part A
What is the radius of the hydrogen-atom Bohr orbit shown in the figure? (Figure 1)
r = ____ nm
The radius of the hydrogen-atom Bohr orbit shown in the figure is 5.3 nm.
What is Bohr orbit?The path that hypothetical electrons take around the nucleus is known as Bohr's orbit.
These orbits are described by Bohr in his hypothesis of the structure of an atom as energy levels or shells where electrons move in a fixed circle around the nucleus.
These orbits resemble solar system orbits, with the exception that they are attracted by electrical forces rather than gravity. The term "ground state" refers to the amount of energy that an electron typically occupies.
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8. A 2kg object explodes and divides into three pieces, one piece has a mass of 800g and has a velocity of [30] m/s, a second piece has a mass of 500g and has a velocity of [520] m/s. What is the velocity of the third mass?
Answer:
v3 = 0 gm/s / 700g
Explanation:
To solve this problem, you need to use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the total momentum of the system (the 2kg object before it explodes) is equal to the sum of the momenta of the three pieces after the explosion.
You can calculate the momentum of each piece by multiplying its mass by its velocity:
P1 = 800g * 30 m/s = 24,000 gm/s
P2 = 500g * 520 m/s = 260,000 gm/s
The total momentum of the system is the sum of these two momenta:
Ptotal = P1 + P2 = 24,000 gm/s + 260,000 gm/s = 284,000 g*m/s
The third piece has a mass of 2kg - 800g - 500g = 700g. We can use the conservation of momentum equation to find its velocity:
Ptotal = (700g * v3) + (800g * 30 m/s) + (500g * 520 m/s)
v3 = (Ptotal - (800g * 30 m/s) - (500g * 520 m/s)) / 700g
v3 = (284,000 gm/s - (800g * 30 m/s) - (500g * 520 m/s)) / 700g
v3 = (284,000 gm/s - 24,000 gm/s - 260,000 gm/s) / 700g
v3 = (284,000 - 24,000 - 260,000) gm/s / 700g
v3 = 0 gm/s / 700g
The velocity of the third mass is 0 m/s.
Hope this helps.
with what speed will a clock have to be moving in order to run at a rest that is one half the rate of clock at rest
The speed at which a clock would have to be moving in order to run at half the rate of a clock at rest depends on the theory of relativity that you are using.
What is the clock speed about?In special relativity, time dilation is the phenomenon where time appears to pass differently for objects in motion relative to an observer at rest.
According to the theory, time appears to slow down for an object as it approaches the speed of light. The rate at which time appears to pass for an object is given by the equation:
T' = T / [tex]\sqrt{(1 - (v^2 / c^2))}[/tex]
Where T is the time as measured by an observer at rest, T' is the time as measured by an observer moving relative to the object, v is the velocity of the object, and c is the speed of light.
In addition, this is a theoretical scenario, practically in order to measure time dilation in a laboratory, it is required a very high precision of measurements, that are currently not possible.
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A 4.51 kg object is placed upon an inclined plane which has an incline angle of 23.0*. The object slides down the inclined plane with a constant speed. Find the normal force, friction force and the coefficient of sliding friction
To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).
How to find the normal force ?The weight of the object is (4.51 kg) * (9.8 m/s^2) = 44.398 NTo find the friction force, we can use the equation: friction force = coefficient of friction * normal force.We can assume that the friction force is equal to the force of gravity acting against the object because it is moving down the inclined plane at a constant pace. As a result, the friction force is equal to the product of the object's weight and sin (incline angle)Friction force is equal to (9.927 N)*sin(23.0)*(44.398 N)We can use the following equation to determine the coefficient of sliding friction:friction coefficient is calculated as friction force divided by normal force.coefficient of sliding friction = 9.927 N /44.398 N = 0.224Therefore, the normal force is 44.398 N, the friction force is 9.927 N, and the coefficient of sliding friction is 0.224.To know more about normal force , check out :
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Grandma Sue (mass 80 kg) and her grandson James (mass 40 kg) are on a smooth icy surface. As Grandma Sue whizzes around the icy surface at 3 m/s in a straight line, she is suddenly confronted with scared James standing at rest directly in her path. Rather than knock him over, she picks him up and continues her uniform motion in a straight line without braking. Find the speed of Grandma Sue and James after the collision.
Answer:
v = 2 m/s
Explanation:
Here, we will use the law of conservation of momentum to solve this problem:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of grandma = 80 kg
m₂ = mass of James = 40 kg
u₁ = initial speed of grandma = 3 m/s
u₂ = initial speed of James = 0 m/s
v₁ = v₂ = v = final speed of grandm and James = ?
Therefore,
[tex](80\ kg)(3\ m/s)+(40\ kg)(0\ m/s)=(80\ kg)(v)+(40\ kg)(v)\\\\(120\ kg)v = 240\ Ns\\\\v = \frac{240\ N.s}{120\ kg}\\[/tex]
v = 2 m/s
You apply a net force on a soccer ball of 15 N. If the acceleration it has is 5 m/s2 what is the mass of the ball?
Answer:
3 kgExplanation:
The mass of the ball can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
We have
[tex]m = \frac{15}{5} = 3 \\ [/tex]
We have the final answer as
3 kgHope this helps you
QUESTION 4
A student lifts a 400 N sandbag 2 meters off the ground. How much work, in joules, did the student perform?
Answer:
800J
Explanation:
W = Fs, Work equals force times displacement
in this case, the force is 400N and the displacement is 2 meters.
The regular SI unit for work is joules
three way to calculate average are
Which of the following is the only group in mission control that gets to communicate with the astronauts in space?
Flight Director
CAPCOM
Senior Flight Controller
Lead Ground Astronaut
Capsule communicator or Capcom is the only group in mission control that gets to communicate with the astronauts in space. Hence, option (B) is correct.
What is capsule communicator or Capcom?The capsule communicator, or Capcom, was the only voice that spoke to the astronauts during their trip to avoid any mistake. To ensure that the men in the capsule always had a familiar individual who understood their situation and could provide the information they required, Capcom was always manned by astronauts.
Both the technical control team on the ground and the astronauts in space are represented by Capcom.
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Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which radiates with a peak wavelength of about 970 nm? K (b) Rigel, a bluish-white star in Orion, radiates with a peak wavelength of 145 nm. Find the temperature of Rigel's surface. K
Answer:
(a) T = 2987.6 k
(b) T = 19986.2 k
Explanation:
The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:
[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}[/tex]
where,
T = Radiated surface temperature
[tex]\lambda_{max}[/tex] = peak wavelength
(a)
here,
[tex]\lambda_{max}[/tex] = 970 nm = 9.7 x 10⁻⁷ m
Therefore,
[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}[/tex]
T = 2987.6 k
(b)
here,
[tex]\lambda_{max}[/tex] = 145 nm = 1.45 x 10⁻⁷ m
Therefore,
[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}[/tex]
T = 19986.2 k
examine the following graph.
a) What is the amplitude of the oscillation?
b) What is the period of the oscillation?
A hair dryer uses 1200 watts of power. Current flow through
the dryer is 10 amperes. At what potential difference does the hair dryer operate
Answer:
did any of this help
Explanation:
y = (-2/3)x - 1
y-(-5)= -2/3(x-6)
y-y1=m(x-x1)
2x-3y=11
Some bat species have auditory systems that work best over a narrow range of frequencies. To account for this, the bats adjust the sound frequencies they emit so that the returning, Doppler-shifted sound pulse is in the correct frequency range. As a bat increases its forward speed, should it increase or decrease the frequency of the emitted pulses to compensate?
Answer:
As a bat increases its forward speed, it should decrease the frequency of the emitted pulses to compensate.
decrease
Explanation:
Decreasing the frequency of the emitted pulse will help the bat reduce its frequency caused by its forward motion. The forward motion shifts the bat's auditory frequency to a higher frequency; consequently, the bat should adjust downwards the frequency of the emitted pulse so the reflected pulse will be in the correct frequency range.
3. Which of the forms of electromagnetic radiation listed below has the greatest energy?
Ogamma rays
O infrared
O ultraviolet
radio waves
Answer:
Choice A. gamma rays
Explanation:
In each of the four situations below an object is experiencing (nearly) uniform circular motion. State what force is providing the centripetal force required to keep the object moving in a circle: a. A car driving around a track. b. A ball being swung on the end of a string. c. The moon orbiting the earth. d. A rotating wheel.
Answer:
Explanation:
Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.
Hence;
1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.
2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.
3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.
4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.
The force that provides the centripetal force in each of the given situations are;
A) Friction Force
A) Friction ForceB) Tension Force
A) Friction ForceB) Tension ForceC) Force of gravity
A) Friction ForceB) Tension ForceC) Force of gravity D) No centripetal force
When an object is in circular motion, the force that keeps it moving round the circle while centrifugal force is the one that tries to pull the object away from the center.
A) When a car is driving around a track, there is a frictional force between the tires of the car and the track that acts on the vehicle to keep it in that circular motion. This frictional force is the centripetal force required to keep the vehicle in circular motion.
B) When a ball is swing on the end of a string, there is an upward force called tension force that acts on the ball to keep it swinging in circular motion. Thus, the centripetal force here is provided by the tension force.
C) When the moon is orbiting the earth, there is a force of gravity exerted by the earth on the moon that keeps the moon in a circular motion about Earth instead of moving in a straight line.
D) For a rotating wheel, the centripetal force does not do any work. The reason for that is because the centripetal force points toward the center of the circle, and as a result it means that the velocity of the rotating wheel is tangent to the circle.
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