Anybody know the answer to this ?

Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03

Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

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8 0

2 years ago

Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .

Vlad [161]

<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

= 4 - (-5)

= 9 cm

Thus;

work done = 95 × 9/100

<u>= 8.55 Joules </u>

5 0

2 years ago

After turning off the television, you approach it. As you get close, but not touching it, the hairs on your arm start to stand u

Amanda [17]

The answer is D) induction.

6 0

2 years ago

Read 2 more answers

A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.A

kenny6666 [7]

Answer: 1.76 Nm

Explanation:

If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.

In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:

τ = F * r = 16 N. (0.11) m = 1.76 Nm

7 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:

$\Delta U = mg \Delta h$