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matrenka [14]
3 years ago
13

When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500

N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 760 N , including the "friction" of turning the truck's drivetrain. What is the magnitude of the force the car applies to the truck?
Physics
1 answer:
likoan [24]3 years ago
5 0

Answer:

3740N

Explanation:

The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object as shown in the formula below. If the net force acting on an object is zero, then the object is not accelerating and is in a state that we call equilibrium.

The formula for force is force = mass x acceleration.

To find mass, simply divide the force by the acceleration

acceleration=1.1m/s²

Mass=4500−760=3400N

Therefore, Mass x acceleration

=1.1m/s²x3400=3740N

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2 years ago
Which of the the following distance vs time graphs represents an object the is moving at constant non zero velocity
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A graph with a horizontal line
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3 years ago
A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su
madreJ [45]

maximum static friction acting on the object will be

F_s = \mu_s mg

plug in all values

F_s = 0.40 \times 15 \times 9.8 = 58.8 N

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0
3 years 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

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From her bedroom window a girl drops a water-filled balloon to the ground, 4.60 m below. If the balloon is released from rest, h
anygoal [31]
Known variables
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d=1/2gt2
4.6=1/2 -9.8 t2
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3 years ago
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