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alexandr1967 [171]
2 years ago
13

When is a train on Earth in motion relative to the sun? A It is relative to the sun only when the train is not moving. B It is n

ever in motion relative to the sun. C It is relative to the sun only when the train is moving. D It is always in motion relative to the sun.
Physics
1 answer:
grandymaker [24]2 years ago
7 0

Answer:

A. It is relative to the sun only when the train is not moving.

Explanation:

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A bicycle travels 22 km in 0.50 hr. What is its average speed?
algol [13]

Answer:

22/0.50=44

Explanation:

average speed= distance / time

8 0
3 years ago
A 3,000 kg truck moving at +10 m/s hits a 1,000 kg parked car which moves off at +15 m/s What is the velocity of
Rina8888 [55]

Answer:

v₃ = 5 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

P = m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision

Pbeforecollision = Paftercollision

(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the truck = 3000 [kg]

v₁ = velocity of the truck = 10 [m/s]

m₂ = mass of the car = 1000 [kg]

v₂ = velocity of the car before the collision = 0 (the car is parked)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 15 [m/s]

(3000*10) + (1000*0) = (3000*v₃) + (1000*15)

30000 = 3000*v₃ + 15000

3000*v₃ = 30000 - 15000

3000*v₃ = 15000

v₃ = 5 [m/s]

7 0
2 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

8 0
3 years ago
If humans are the most complex creature, does that mean they have the most number of chromosomes?
Bezzdna [24]
Yes. Either way though, humans have more chromosomes than any other species.

5 0
3 years ago
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4 0
2 years ago
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