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Delicious77 [7]
3 years ago
11

Apply the law of conservation of energy and describe the energy transformations that occur as you coast down a long hill on a bi

cycle and then apply the brakes to make the bike stop at the bottom.
A roller coaster is at the top of a hill and rolls to the top of a lower hill. If mechanical energy is constant, then on the top of which hill is the kinetic energy from the roller coaster`s motion greater?
Physics
1 answer:
m_a_m_a [10]3 years ago
4 0
As you coast down a long hill on your bicycle, potential energy from your height is converted to kinetic energy as you and your bike are pulled downward by gravity along the slope of the hill. While there is air resistance and friction slowing you down by a little bit, your speed increases gradually until you apply the brakes, causing enough friction to slow yourself and the bike to a stop at the bottom.

A roller coaster will have higher kinetic energy at the lower hill because it will have already been moving as opposed to the initial hill. But I'm not one hundred percent certain. You can always google this stuff, but I do know for sure that at the first hill, the roller coaster will have higher potential energy.
Hope this helps!
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A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.
OLga [1]

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

  • <em>Mass of the block, m = 2.5 kg</em>
  • <em>Spring constant, k = 56 N/m</em>
  • <em>Extension of the spring, x = 0.75 m</em>

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5  \ m/s

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here:  brainly.com/question/166559

3 0
2 years ago
Read 2 more answers
Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook
jeka57 [31]

Answer: a) Mr = 2.4×10^-4kg/s

V = 34.42m/a

b) E = 173J

Ø = 2693.1J

c) Er = 0.64J/s

Explanation: Please find the attached file for the solution

3 0
4 years ago
If a frog jumps with an initial velocity of 100m/s at an angle of 60°. a. What is the frog’s x and y components of velocity? b.
Nadya [2.5K]

Explanation:

It is given that,

Initial velocity of frog, u = 100 m/s

It jumps at an angle of 60 degrees.

(a) Let v_x\ and\ v_y are the x and y components of velocity. It can be calculated as :

u_x=u\ cos\theta

u_x=100\times \ cos(60)

u_x=50\ m/s

u_y=u\ sin\theta

u_y=100\times \ sin(60)

u_y=86.6\ m/s

(b) Let y is the maximum height reached by the frog. It can be calculated using the third equation of motion as :

At maximum height, v_y=0

v_y^2-u_y^2=2ay and a = -g

-u_y^2=-2g

y=\dfrac{u_y^2}{2g}

y=\dfrac{86.6^2}{2\times 9.8}

y = 382.63 meters

(c) Let t is the time taken by the frog to reach its maximum height. It can be calculated as :

v_y=u_y-gt

0=u_y-gt

t=\dfrac{u_y}{g}

t=\dfrac{86.6\ m/s}{9.8\ m/s^2}

t = 8.83 seconds

Hence, this is the required solution.

8 0
3 years ago
How does car brakes change physical forces related to:
professor190 [17]
Acceleration and deceleration and momentom 
they slow the car down and it loses momentome

8 0
3 years ago
In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a
pickupchik [31]

Answer:

time of collision is

t = 0.395 s

h = 5.63 m

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

v_1 = 0

similarly initial speed of the object which is projected by spring is given as

v_2 = 16.2 m/s

now relative velocity of object with respect to ball

v_r = 16.2 m/s

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

d = v_r t

6.4 = 16.2 t

t = 0.395 m

Now the height attained by the object in the same time is given as

h = v_2 t - \frac{1}{2}gt^2

h = 16.2(0.395) - \frac{1}{2}(9.81).395^2

h = 5.63 m

so they will collide at height of 5.63 m from ground

4 0
4 years ago
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