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Ira Lisetskai [31]
3 years ago
14

Two rocks of different masses are rolling down a hill at the same speed. Which rock (small one or big one) would have more kinet

ic energy?
Physics
2 answers:
Angelina_Jolie [31]3 years ago
6 0

Answer:

The larger rock would roll down the hill faster

Explanation:

aivan3 [116]3 years ago
3 0

Answer:

Kinetic energy of bigger rock will be more than that of smaller one.

Explanation:

Kinetic energy of the rock is given by,

Kinetic energy = \frac{1}{2} m v^{2}

As velocity of both the rocks are same. Thus, kinetic energy is directly proportional to the mass of the rock

Kinetic energy ∝ mass

So, For greater mass kinetic energy will be greater and for smaller mass kinetic energy will be smaller.

Hence, Kinetic energy of bigger rock will be more than that of smaller one.

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Pls answer quick will mark brainlest and 5 stars
QveST [7]

Answer:A

Explanation:

It’s bigger I am not sure

7 0
3 years ago
Read 2 more answers
A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse
eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5  

=> n2 = sin51 / sin19.5

 = 2.33

7 0
3 years ago
Show that the Mass spring system executes simple harmonic motion(SHM)?​
murzikaleks [220]

Explanation:

Show that the motion of a mass attached to the end of a spring is SHM

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed

at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.

If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

According to "Hook's Law

F = - Kx ---- (1)

Negative sign indicates that the elastic restoring force is opposite to the displacement.

Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion

from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by

F = ma ---- (2)

Comparing equation (1) & (2)

ma = -kx

Here k/m is constant term, therefore ,

a = - (Constant)x

or

a a -x

This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.

5 0
3 years ago
A car is traveling south is 200 kilometers from it’s starting point after 2 hours. What is the average velocity of the car
Lesechka [4]

Answer:

100

Explanation:

take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time) = 100

6 0
3 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
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