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Andrews [41]
4 years ago
5

A rock falls from the top of a hill to the bottom. The total energy of the rock at the top isn’t the same as the sum of its pote

ntial energy and kinetic energy at the bottom which state meant best explains this situation
A) some of the rocks energy is transformed to thermal energy
B) the rolling rock loses some of its energy because of gravity
C) the rolling rock doesn’t follow the law of conservation of energy
D) the rock absorbs some of its own potential energy
Physics
1 answer:
mart [117]4 years ago
3 0

Answer:

A) some of the rocks energy is transformed to thermal energy

Explanation:

If we neglect air resistance during the fall of the rock, than the mechanical energy of the rock (which is sum of its potential energy and its kinetic energy) would be constant during the entire motion, so the total energy of the rock at the top would be the same as the sum of its potential energy and kinetic energy at the bottom.

However, this not occurs, due to the presence of air resistance. In fact, air resistance acts against the fall of the rock, and because of the friction between the molecules of air and the surface of the rock, the rock loses part of its energy. This energy is converted into thermal energy of the molecules of the air.

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It has to be one continuous column of cloud (air) connected to the ground and in constant rotation.
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Help with physics projectile motion​
BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0
3 years ago
The force that pulls planets torwards the sun is called​
Studentka2010 [4]

Answer:

gravity

Explanation:

Gravity pulls the planets toward the Sun. Gravity pulls the moon toward Earth. Gravity pulls us toward the Earth. Gravity is a force. Inertia.

7 0
3 years ago
A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

           v_y = v_{oy} - g t

           v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)

at the highest point of the trajectory the vertical speed must be zero

           v_y = 0

therefore the velocity of the body is

          v = (10 i ^ + 0j ^) m / s

the acceleration is

          a = (0 i ^ - g j⁾

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5 0
3 years ago
What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
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