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bagirrra123 [75]
2 years ago
8

How do kinetic and potential energy transfer to one throughout a roller coaster ride?

Physics
1 answer:
mojhsa [17]2 years ago
6 0

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

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A device has power -2 D . The device is
slavikrds [6]

Answer:

b.

Explanation:

-vesign shows the lens is <em><u>C</u></em><em><u>O</u></em><em><u>N</u></em><em><u>C</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em>

<em><u>f</u></em><em><u>=</u></em>1/power

7 0
3 years ago
How could you increase the force advantage of a lever?Select one:a. Make the effort length longer.b. Make the effort length shor
11Alexandr11 [23.1K]

The ideal mechanical advantage of a lever (IMA) is given by:

IMA=\frac{Le}{Lr}

Where:

Le = Effort of the arm

Lr = Resistance arm.

Therefore, we can increase the force adventage by increasing the effort arm or reducing the load arm

Answer:

a. Make the effort length longer.

4 0
1 year ago
The temperature of the cosmic background radiation is measured to be 2.7 k. What is the wavelength of the peak in the spectral d
KATRIN_1 [288]

Answer:

1.07\cdot 10^{-3} m

Explanation:

The peak wavelength of the spectral distribution can be found by using Wien's displacement law:

\lambda=\frac{b}{T}

where

b=2.898\cdot 10^{-3} m\cdot K is Wien's displacement constant

T is the absolute temperature

For the cosmic background radiation, the temperature is

T = 2.7 K

So, the corresponding peak wavelength is

\lambda=\frac{2.898\cdot 10^{-3} m\cdot K}{2.7 K}=1.07\cdot 10^{-3} m

7 0
3 years ago
A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an
Roman55 [17]

Answer:

v_{2f} = \frac{2vm_1}{m_2 + m_1}

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

P_i = m_1v

After the collision

P_f = m_1v_{1f} + m_2v_{2f}

So using the law of momentum conservation

P_i = P_f

m_1v = m_1v_{1f} + m_2v_{2f}

We can solve for the speed of ball 1 post collision in terms of others:

v_{1f} = v - v_{2f}\frac{m_2}{m_1}

Their kinetic energy is also conserved before and after collision

m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2

m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2

From here we can plug in v_{1f} = v - v_{2f}\frac{m_2}{m_1}

m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2

m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2

m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2

v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0

v_{2f}(1 + \frac{m_2}{m_1}) = 2v

v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}

8 0
3 years ago
Read 2 more answers
A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s, and the stone is 1.50 m above the ground when lau
arlik [135]

Answer: a) 19.21m b) 3.92secs

Explanation:

a) Maximum height reached by the object is the height reached by an object before falling freely under gravity.

Maximum height = U²/2g

U is the initial velocity = 19.6m/s

g is acceleration due to gravity = 10m/s²

Maximum Height = 19.6²/2(10)

H = 19.21m

b) The time elapsed before the stone hits the ground is the time of flight T= 2U/g

T= 2(19.6)/10

T = 39.2/10

Time elapsed is 3.92secs

5 0
3 years ago
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