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tensa zangetsu [6.8K]

3 years ago

10

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

s the best approximation of the box’s acceleration? 120 m/s2 20 m/s2 1 m/s2 9 m/s2

1 answer:

Hoochie [10]3 years ago

8 0

The acceleration of the box is approximately $1 m/s^2$

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

$\sum F = F_a - F_f$

where

$F_a = 20 N$ is the applied forward force

$F_f=9.0 N$ is the friction force

Combining the two equations,

$F_a-F_f=ma$

And solving for the acceleration,

$a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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3 years ago

A petrol tanker ha 2800kg when empty and hold 30m3 0f petrol when full. The denity of petrol i 740kg/m3. Calculate the ma of the

White raven [17]

The mass of the tanker with petrol is 250000 N.

We are given that,

Mass of tanker= m =2800 kg

Volume of petrol= v =30 m³

Density of petrol= d =740 kg/m³

Thus , mass , density and volume relation can be given as,

density= mass/ Volume

Mass = Density × Volume

Mass = 740× 30

Mass = 22200 kg

The mass of the petrol is 22200 kg.

Total mass of tanker with petrol = Mass of petrol + Mass of tanker

Total mass of tanker with petrol= 22200+ 2800 kg

Total mass of tanker with petrol= 25000 kg

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Where, weight = m × g ,(g =10m/s²)

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1 year ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago