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LuckyWell [14K]
3 years ago
14

Look at the four positions of Earth with respect to the sun.

Physics
1 answer:
crimeas [40]3 years ago
4 0

Answer:

position 3

Explanation: HOPE IT HELPED

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Calculate The pressure produced by a force of 392 N acting on an area of 8.0 m^2
Nonamiya [84]

Answer:

Explanation:

500-0.05=499.95

500-100+5=405

100+100=10000

5 0
2 years ago
3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) movin
Alex787 [66]

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

6 0
3 years ago
A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a
GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

7 0
3 years ago
How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^
arlik [135]

Answer:

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^2)?

Explanation:

<em>1700N </em>

<em> Mass multiplied by acceleration gives you the amount of force needed for it.</em>

3 0
2 years ago
Read 2 more answers
Operators must slow to no wake speed when within what distance of a u.s. naval vessel?
Alex Ar [27]
Recreational boaters have a role in keeping the waterways safe and secure. Violators of the restrictions can expect a quick and severe response. The operator should not approach within yards and slow to minimum speed within 500 yards of any of united states naval vessel. If the vessel need to pass within 100 yards of a united states naval vessel for safe passage, must contact the united states naval vessel or the united states coast guard escort vessel on VHF-FM channel 16. In addition,

• The operator must perceive and avoid all security areas.

• Avoid commercial port operation areas particularly those that include military cruise line or petroleum amenities. 

• The operator must perceive and avoid other controlled areas near dams, power plants, etc. 

• The operator of the vessel must not discontinue or anchor underneath bridges or in the channel. 
6 0
3 years ago
Read 2 more answers
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