100 POINTS WILL GIVE BRAINIEST TO BEST ANSWER!!!!!!!!!!!!!<br><br> see below

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

What do scientists hope to learn from missions to visit asteroids?

Lady_Fox [76]

Answer:

The mission will help scientists investigate how planets formed and how life began, as well as improve our understanding of asteroids that could impact Earth.

Explanation:

Hope this helps :)

8 0

2 years ago

how much time would it take for the sound of a thunder to travel 1,500 meters if sound travelers at a speed of 330 m/sec?

inysia [295]

(1,500 meters) x (1 sec/330 meters) =

(1,500 / 330) (meters-sec/meters) =

4.55 seconds

8 0

3 years ago

During the 1840s, Americans used the idea of Manifest Destiny to justify the...

Natali [406]

B because was the idea that Americans were made to civilize the American territory. That ideia was created by God, i mean, the destiny of america was chosen by god.

3 0

2 years ago

Read 2 more answers

Yvette hangs a 2.4kg bird feeder in the middle of a rope tied between two trees, the feeder creates a

kicyunya [14]

Answer:

a) 0.049 m

b) Yes, increase

Explanation:

Draw a free body diagram.

In the y direction, there are three forces acting on the feeder. Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.

Apply Newton's second law to the feeder in the y direction.

∑F = ma

2Ty − mg = 0

Ty = mg/2

Let's say the distance the rope sags is d. The trees are 4m apart, so the feeder is 2m horizontally from either tree. Using Pythagorean theorem, we can find the length of the rope on either side:

L² = 2² + d²

L = √(4 + d²)

Using similar triangles, we can write a proportion using the forces and distances.

Ty / T = d / L

Substitute:

(mg/2) / T = d / √(4 + d²)

Solve for d:

Td = mg/2 √(4 + d²)

T² d² = (mg/2)² (4 + d²)

T² d² = (mg)² + (mg/2)² d²

(T² − (mg/2)²) d² = (mg)²

d² = (mg)² / (T² − (mg/2)²)

d = mg / √(T² − (mg/2)²)

Given m = 2.4 kg and T = 480 N:

d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)

d = 0.049 m

b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.

8 0

2 years ago

100 POINTS WILL GIVE BRAINIEST TO BEST ANSWER!!!!!!!!!!!!! see below