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Sliva [168]
3 years ago
10

An elevator accelerates upward at 2.0 m/s?,

Physics
1 answer:
amm18123 years ago
4 0

Answer:

Force the floor exerts on the  passenger is 833 N.

Explanation:

  • Weight of passenger (F_{g}) = mg = 85 × 9.8 N = 833 N
  • Force the floor exerts on the  passenger (F_{N}) = ?
  • For the elevator with the speed as 2.0 m/s the net force is zero, it means that the force is balanced.

                                       i.e. F_{N} = - F_{g} = -mg = 833 N

                                       hence  F_{N} is 833 N

  • If the lift was not at a constant speed i.e. if it had acceleration (m/s^{2})  then the case would be different.

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A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco
oee [108]
S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

The depth is approximately 78 meters.
(My brain hurts now) :P Good Luck!
4 0
3 years ago
Balanced forces can change an<br> object's direction.<br> A. True<br> B. False
makvit [3.9K]

Answer:

False

Explanation:

Balanced forces result in a net force of 0N. This means no direction or acceleration change will be applied to the object. A torque may be applied, but with no other external forces, the object will not move.

8 0
3 years ago
Railroad car A, with mass 6275 kg, is traveling east at 6.5 m/s along a straight track. It strikes railroad car B, with mass 515
LekaFEV [45]

The speed of A and B immediately after collision is 5.28m/s

<u>Explanation:</u>

Mass of A is 6275kg

Speed of A is 6.5m/s

Mass of B is 5155kg

Speed of B is 3.8m/s

Track is frictionless.

A and B stick together.

speed of attached A and B = ?

mₐsₐ + mᵇsᵇ = (mₐ + mb) s

6275 X 6.5 + 5155 X 3.8 = ( 6275 + 5155) X s\\\\s = \frac{40787.5 + 19589}{11430}\\ \\s = \frac{60376.5}{11430}\\ \\s = 5.28m/s

Therefore, The speed of A and B immediately after collision is 5.28m/s

4 0
3 years ago
When the cart moves on an incline at constant speed, it is in equilibrium; i.e., the net force on it is zero. Does it require mo
Amanda [17]

Answer:

It requires more tension to pull up the track

Explanation:

Net force must be zero to maintain constant velocity.

Weight force will always be pointed down the slope. Call it W

Friction force (Call it Ff) will be down slope when movement is up slope.

Friction force will be up slope when movement is down slope.

W and Ff are always positive numbers

Call the pulling force T

If Up slope is considered the positive direction

Moving up slope

Tu - Ff - W = 0

Tu = Ff + W

Moving down slope

Td + W - Ff = 0

Td = Ff - W

Ff + W > Ff - W therefore Tu > Td

5 0
2 years ago
What is the velocity of the boat if the passenger shoots at an angle 60 with horizontal ? The mass of boat with the passenger is
mixas84 [53]

Answer:

400cos(60) = 200m/s

100(v) = 0.035(200)

v= 0.07m/s

8 0
2 years ago
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