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3 years ago

A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?

Physics

1 answer:

ELEN [110]3 years ago

5 0

Answer:

$\boxed {\boxed {\sf 50 \ kg}}$

Explanation:

We are asked to find the mass of a cabinet, given the force and acceleration. According to Newton's Second Law of Motion, force is the product of mass and acceleration. The formula for this is:

$F= m \times a$

The force is 200 Newtons, but we should convert the units to make unit cancellation easier. 1 Newton is equal to 1 kilogram meter per second squared, so the force of 200 Newtons is 200 kilogram meters per second squared.

The mass is unknown and the acceleration is 4 meters per second per second or 4 meters per second squared.

F= 200 kg*m/s²
a= 4 m/s²

Substitute the values into the formula.

$200 \ kg *m/s^2 = m \times 4 \ m/s^2$

We are solving for the mass, m, so we must isolate the variable. It is being multiplied by 4 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 4 m/s²

$\frac {200 \ kg *m/s^2}{4 \ m/s^2}= \frac{m \times 4\ m/s^2}{4 \ m/s^2}$

$\frac {200 \ kg *m/s^2}{4 \ m/s^2} =m$

The units of meters per second squared cancel.

$\frac {200 \ kg }{4 }=m$

$50 \ kg =m$

The mass of the cabinet is <u>50 kilograms.</u>

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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range

Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

$v_y(t)=vo*sin(A)-g*t$

the velocity is Zero when the projectile reach in the maximum altitude:

$0=vo-gt\\t=\frac{vo}{g}$

When the time is vo/g the projectile are in the middle of the range.

$d_x(t)=vo*cos(A)*t\\$

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$R=d_x(t=2*\frac{vo}{g})$

$R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}$

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

$\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\$

2B=60°

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3 years ago

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

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2 years ago

Please help

inysia [295]

Explanation:

Start with what you know and list your knowns and unknowns

F = ma

F= 3N

m = 6kg

a =?

3N = 6kg x a

solve for a

3N / 6kg = a

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3 years ago

What makes output energy less than input energy?

AleksAgata [21]

Answer:

part of input energy is wasted

it is used to increase the entropy of the surrounding

therefore the useful energy output is necessarily smaller than the energy input

in other words the efficiency of heat engine is always less than hundred percent

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2 years ago

Read 2 more answers

A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?​

A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?