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frosja888 [35]
3 years ago
13

Peter and Fred are skateboarding in a large semicircular halfpipe. Peter starts out from rest at a height h and collides with Fr

ed standing at the bottom. Peter and Fred have about the same mass. After the collision, which height will Fred reach?
Physics
1 answer:
jolli1 [7]3 years ago
6 0

Answer:

The answer is h/4

Explanation:

When Peter collides with Fred, the collision is inelastic & they both proceed with a velocity of V/2.

let m represent the masses for Peter & Fred

v represent the initial velocity of Peter

V represent final velocity of both of them

mv + 3m × 0 = (m+m)V

V = v/2

Using the expression; H = v² / 2g .............Eqn 1

Upon substitution of V/2 into Eqn 1 above,

H = (V/2)² / 2g

H = (V²/4) ÷ 2g

Therefore height will be h/4

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Match each key term with the correct definition.
oksano4ka [1.4K]

Answer:

Radioisotope -> An atom with an unstable nucleus

Radioactivity -> The spontaneous discharge of energy from an unstable nucleus

Radioactive Decay -> The process by which the nucleus of an unstable isotope changes

Strong Nuclear Force -> Binds protons and neutrons together in the nucleus

Explanation:

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3 years ago
a child rides her bicycle in her level drive way. what form of energy does she possess A. Elastic B. Nuclear C. Potential D. Kin
Bad White [126]

Answer:

Kinetic Energy

Explanation:

Kinetic energy is energy due to motion.

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3 years ago
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A circular test track for cars has a circumference of 4.7 km. A car travels around the track from the southernmost point to the
Alika [10]
Well, for the distance traveled, the car goes from the northernmost point to the southernmost point. So, it travels half of the circle's circumference = 4.7/2 = 2.35 km.

For the displacement, by going from the northernmost point to the southernmost point, the car basically just travels the diameter of circle.

So, using the formula: Circumference = 2πr = <span>πd

Hence, the d = C/</span>π = 4.7/<span>π = 1.49605... = 1.5 km (2 significant figures)
Therefore, displacement = 1.5 km</span>
7 0
3 years ago
PLZ HELP ASAP! TYSM! I WILL AWARD BRAINLIEST!
Taya2010 [7]

Answer:

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Explanation:

7 0
3 years ago
Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th
lozanna [386]

Answer:

Part a)

Velocity of sled

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

Velocity of sled

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

0 = m_1(v - s) + (m_2 + M)v

v = \frac{m_1 s}{m_1 + m_2 + M}

so velocity of the sled + other person is

v = \frac{m_1 s}{m_1 + m_2 + M}

velocity of first man who jump off

v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s

v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f

(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f

Now we have

v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s

Also the speed of second person is given as

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s

v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}

Part c)

change in kinetic energy of sled + two people is given as

KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

here we know all values of speed as we found it in part a) and part b)

4 0
3 years ago
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