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2 years ago

8

Sam is riding his bike and spots a sinkhole. He slows his bike from 14.0 m/s to 2.0 m/s on 6.0 seconds. What is the bikes negati

ve acceleration?

2 answers:

Paul [167]2 years ago

7 0

The bikes negative acceleration is -2(m/s^2)

VARVARA [1.3K]2 years ago

4 0

By definition, average acceleration will be:
a=(v(f)-v(i))/t
a=(2-14)/6=-2 (m/s^2)
________
-2(m/s^2)

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This is a machine that converts electrical energy into mechanical energy.

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<h2>Answer:</h2>

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<h2>Explanation:</h2>

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3 years ago

Multi celled organisms have a ______ surface area to volume ratio than single celled organisms.

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A. Larger. It is larger Bc they r all larger than the other

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2 years ago

un futbolista patea una pelota que se encuentra en el pasto con un angulo de 30° (medido desde la horizontal) con la intención d

Rus_ich [418]

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3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

2 years ago

4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?

True [87]

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

509 J = 12x10xH

509 J = 120xH

H = 509/120

H = 4.24 m

Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.

Thank you

8 0

2 years ago