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3 years ago

Definition: - When a body is moving in circular path at a distance r from its center, this is its velocity at any instant?

Physics

2 answers:

Natali [406]3 years ago

4 0

Answer:

tangential velocity

Explanation:

kupik [55]3 years ago

3 0

Tangential velocity!

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Why is scalar quantity negative charges is not possible

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Answer:

Scalar quantity can never be Negative. Because scalar has only magnitude not direction. And magnitude can't be negative.

Explanation:

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3 years ago

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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)

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3 years ago

True or False?

WINSTONCH [101]

Answer:

true is the answer of the question

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3 years ago

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A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s

zmey [24]

Answer:

The angle is 65.6°.

Explanation:

Given that,

Speed = 2.20 m/s

Distance from the shore= 500 m

Distance from the bottom= 1100 m

Speed of boat = 7.30 m/s

According to figure,

We need to calculate the angle with shore

Using formula of angle

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\tan\theta=\dfrac{500}{1100}$

$\theta=\tan^{-1}(\dfrac{500}{1100})$

$\theta=24.4^{\circ}$

We need to calculate the angle

$\alpha=90-\theta$

Put the value into the formula

$\alpha=90-24.4$

$\alpha=65.6^{\circ}$

Hence, The angle is 65.6°.

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