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podryga [215]
3 years ago
15

Mr. Phillips' car is parked on a steep hill with the brakes applied and the engine off. Because of the car's position, it has gr

avitational potential energy. How could this gravitational potential energy be transformed into kinetic energy?
A. He could push it up the hill.


B. He could release the brakes.


C. He could start the engine.


D. all of these
Physics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

The anser is b

Explanation:

You might be interested in
Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volum
SVETLANKA909090 [29]

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

V_A = \frac{1}{2}V_B

Now, from Charle's Law:

\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}

Hence, the correct option is:

<u>A. TA = TB/2.</u>

6 0
3 years ago
A proton is traveling horizontally to the right at 1.8 × 106 m/s. (a) Find the magnitude and direction of the weakest electric f
Mars2501 [29]

Answer:

528398.4375 N/C opposite to the direction of the proton

3.56\times 10^{-8}\ s

288.24609375 N/C in the same direction of the motion of the electron

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity = 1.8\times 10^{6}\ m/s

s = Displacement = 3.2 cm

a = Acceleration

Mass of electron = 9.11\times 10^{-31}\ kg

Mass of electron = 1.67\times 10^{-27}\ kg

q = Charge of particle = 1.6\times 10^{-19}\ C

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(1.8\times 10^6)^2}{2\times 0.032}\\\Rightarrow a=-5.0625\times 10^{13}\ m/s^2

Electric field is given by

E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{1.67\times 10^{-27}\times -5.0625\times 10^{13}}{1.6\times 10^{-19}}\\\Rightarrow E=−528398.4375\ N/C

The electric field is 528398.4375 N/C opposite to the direction of the proton

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-1.8\times 10^6}{-5.0625\times 10^{13}}\\\Rightarrow t=3.56\times 10^{-8}\ s

The time taken is 3.56\times 10^{-8}\ s

E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{9.11\times 10^{-31}\times -5.0625\times 10^{13}}{-1.6\times 10^{-19}}\\\Rightarrow E=288.24609375\ N/C

The electric field is 288.24609375 N/C in the same direction of the motion of the electron

7 0
4 years ago
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find
nikitadnepr [17]

Answer:

a)  x = v₀ₓ t ,  y = v_{oy} t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

       F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis  

       -W = m a_{y}

       -m g = m a_{y}  

       a_{y}  = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

       x = v₀ₓ t + ½ a  t²

As the acceleration on this axis is zero

      x = v₀ₓ t

Y Axis

       y = v_{oy} t + ½ a_{y}  t²

The acceleration on this axis is –g

        y = v_{oy} t - ½ g t²

B) to find the maximum value of distance r

        r =√ x² + y²

        r = √( v₀ₓ² t² + (v_{oy} t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

      dr / dt = 0

      0 = ½ 1/√( v₀ₓ² t² + (v_{oy} t + ½ g t²)²    (v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)(v_{oy} - ½ g2t)

We simplify this expression

         0 = v₀ₓ² 2t + 2 (v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)

       -v₀ₓ² t =( v_{oy} t - ½ g t²)  (v_{oy} - ½ g2t)  

      -v₀ₓ² t = v_{oy}² t -3/2 gt² v_{oy} + 1/2 g² t³  

       ½ g² t²- 3/2 g v_{oy} t  = v_{oy}² + v₀ₓ²

Let's use trigonometry to find go and vox

         sin θ = v_{oy} / v₀

         cos θ = vox / v₀

         v_{oy} = v₀ sin θ

        v₀ₓ = v₀ cos θ

We replace

         ½ g² t² -3/2 g v_{oy} t = v₀ (sin² θ +  cos² θ)

          g t² - 3v₀ sin θ t = 2 v₀/g

 

The time is maximum for the angle is zero

         

8 0
3 years ago
Which is a sub-atomic particle?
Natalka [10]
A particle that is smaller than an atom or a cluster of particles.
4 0
3 years ago
Calculate the pressure of water at the bottom of 6m depth well.( g= 9.8m/s square<br>​
Alexeev081 [22]

Answer:

58.8NM-²

Explanation:

The pressure is the product of density,depth(height) and acceleration due to gravity

here the density of water is 1gm-³ it a constant so it not given

pressure= 6×9.8×1

pressure = 58.8NM-²

8 0
3 years ago
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