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FinnZ [79.3K]
3 years ago
8

A spring with a mass attached is initially at rest. It is then stretched 2 cm by a physics student. Another physics student with

a similar mass and spring system then stretches their spring to 4 cm. Compare and contrast the potential energies of the springs after they are stretched by the students. *
Physics
1 answer:
Andrew [12]3 years ago
3 0
The student who displaced the spring by 2 cm has less potential energy than the student who displaced the spring by 4 cm, this is because potential energy (elastic) is directly proportionate to extension (displaced amount), so as the amount of displacement of the spring is higher, then the potential energy of the springs is higher and vice versa.
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A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri
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The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
                        P = W = mass x acceleration due to gravity
    
                    P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N

Solving for the static friction force (F), 
                                              F = P x c 
 
                                      F = (2.94 N) x 0.6 = 1.794 N

Therefore, the maximum force of static friction is 1.794 N. 



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Explanation:

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Which best describes the transition from gas to liquid? (2 points) Select one: a. Energy must be removed because particles in li
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A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
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Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
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