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3 years ago

The crowbar pivots around a fixed point and makes it easier to pull the nail out. What type of simple machine is the crowbar

Physics

1 answer:

kherson [118]3 years ago

4 0

Answer:

A crowbar is a level.

Explanation:

A crowbar is a first class lever. One where the fulcrum is between the effort and the load; The effort is applied to the end of a crowbar for example pulling up a nail the fulcrum is where the crow bar bends upwards touching the board, the load is the nail on the opposite end of the bar.

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When velocity is graphed with respect to time, what does the area under

Juliette [100K]

Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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3 years ago

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A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85

artcher [175]

Answer:

$v = 10.75\,\frac{m}{s}$

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

$(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )$

The velocity of the bowling ball after the collision is:

$v = 10.75\,\frac{m}{s}$

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3 years ago

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A softball with a mass of 0.11 kg goes moves at a speed of 12 m/s, then the ball is hit by a bat and rebounds in the opposite di

dimulka [17.4K]

=1244 and that's from my math book

7 0

3 years ago

From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft

zhannawk [14.2K]

Answer:

$W=\frac{773}{4.45}=173.76 l b f$

Explanation:

$W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}$

The law of gravitation

$G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)$

Universal gravitational constant [S.I. units]

$m_{e}=5.976\left(10^{24}\right) k g$

Mass of Earth [S.I. units]

$m=89 kg$

Mass of a man in a spacecraft [S.I. units]

$R=6371 \mathrm{~km}$

Earth radius [km]

Distance between man and the earth's surface

$h=261 \mathrm{~km} \quad[\mathrm{~km}]$

ESULT $W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}$

$W=\frac{773}{4.45}=173.76 l b f$

4 0

3 years ago

A 86 kg human stands on the surface of Venus. The mass of Venus is 4.9 × 1024 kg and its radius is 6.1 × 106 m. Collapse questio

iris [78.8K]

Answer:

755.37 N

Explanation:

We are given that

Mass of venus= $m_1=4.9\times 10^{24}kg$

Radius= $r=6.1\times 10^6$ m

Mass of human= $m_2=86 kg$

We know that the gravitational force between two bodies

$F=G\frac{m_1m_2}{r^2}$

Where G=Gravitational constant= $6.67\times 10^{-11}Nm^2/kg^2$

Using the formula

The magnitude of the gravitational force exerted by Venus on the human= $F=\frac{6.67\times 10^{-11}\times 86\times 4.9\times 10^{24}}{(6.1\times 10^6)^2}$

The magnitude of the gravitational force exerted by Venus on the human=F= $755.37N$

7 0

3 years ago