3 years ago

Help please I’ll mark as brainliest

Physics

1 answer:

IceJOKER [234]3 years ago

5 0

Sorry for the messy handwriting but this is what I got

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What has more kinetic energy 15 kg ball rolling north at 15 m/s or a 15 kg ball rolling backwards at 7m/s

Setler79 [48]

Answer:

15 kg ball

Explanation:

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3 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

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4 years ago

The pirate ship tie at the amusement park is a giant pendulum that riders sit in. It swings back and forth, with a maximum veloc

kkurt [141]

Here as we know that there is no loss of energy

so we can say that maximum kinetic energy will become gravitational potential energy at its maximum height

So here we have

$\frac{1}{2}mv^2 = mgh$

here we have

v = 20 m/s

m = 8000 kg

now from above equation we have

$\frac{1}{2}(8000)(20^2) = (8000)(9.8)h$

$h = \frac{200}{9.8)$

$h = 20.4 m$

so maximum height is 20.4 m

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4 years ago

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a girl that has a mass of 55kg is standing on a floor, how much supporting force does the floor have?

KengaRu [80]

A girl standing on a floor would have two opposite forces acting on it. These forces are the weight and the normal force. Since no other forces are acting and that the girl is at rest, then the weight must equate to the normal force. Therefore, the supporting force would be:

F = mg = 55kg (9.81 m/s^2) = 539.55 N

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3 years ago

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I think the second one

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3 years ago