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3 years ago

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The snowblower driver figured that he moved 1250 kg of snow. If the plow pushes with a force of 40000 Newton’s, how fast did the

snow accelerate? (Assume no friction)

1 answer:

alexgriva [62]3 years ago

4 0

The acceleration of snow is 32 m/ $s^{2}$ .

<u>Explanation:</u>

According to Newton's second law of motion, the force acting on any object is directly proportional to the product of mass and acceleration acting on the object.

So if the object is a snowblower driver with mass 1250 kg and the force acting on it is 40000 N, then the acceleration of the snow can be determined from the ratio of force to mass.

Force = Mass × Acceleration

Acceleration = $\frac{Force}{Mass}$ = $\frac{40000}{1250}=32 m/s^{2}$

Thus, the acceleration of the snow is 32 m/ $s^{2}$ .

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raketka [301]

Explanation:

It is given that,

Mass of the man, m = 68 kg

Terminal velocity of the man, v = 59 m/s

We need to find the rate at which the internal energy of the man and of the air around him increase. The gravitational potential energy of the man is given by :

$E=mgh$

Differentiating equation (1) wrt t as :

$\dfrac{dE}{dt}=mg\dfrac{dh}{dt}$

Since, $v=\dfrac{dh}{dt}=59\ m/s$

$\dfrac{dE}{dt}=68\times 9.8\times 59$

$\dfrac{dE}{dt}=39317.6\ J/s$

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3 years ago

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nikklg [1K]

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3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

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Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

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