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3 years ago

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What is the first law of gravity??

1 answer:

neonofarm [45]3 years ago

6 0

Answer:

every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

Explanation:

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Soloha48 [4]

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3 years ago

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what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N

Oksana_A [137]

Answer:

50

Explanation:

The mechanical advantage of a machine is given by

$MA=\frac{F_{out}}{F_{in}}$

where

$F_{out}$ is the output force

$F_{in}$ is the input force

For the crowbar in this problem,

$F_{in}=10 N$ is the force in input applied by the worker

$F_{out}=500 N$ is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

$MA=\frac{500}{10}=50$

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3 years ago

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An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b

trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

$Power = \frac{Work \hspace{1mm} done}{Time}$

$P=\frac{W}{t}$

Work Done, $W= F.s$

Where s is displacement in the direction of force and F is force.

$\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v$

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

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3 years ago

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Arisa [49]

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3 years ago

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A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago