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sergey [27]
1 year ago
13

An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a

cceleration. Is a person in this ride experiencing apparent weightlessness, true weightlessness, or neither? Explain.
Physics
2 answers:
o-na [289]1 year ago
5 0
An object experiences true weightlessness when the net force of all gravitational forces acting upon the object is zero. In this case, the gravitational force exerted by the earth on the people that are on the park ride while it's free falling never ceases to act on the people. If the person on the ride were in a case of true weightlessness then they would not fall in any direction in the first place. The answer is the apparent weightlessness.
blsea [12.9K]1 year ago
3 0

Answer: apparent weighlessness.


Explanation:


1) Balance of forces on a person falling:


i) To answer this question we will deal with the assumption of non-drag force (abscence of air).


ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).


iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).


2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.


3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.


Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

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What are some other examples of scientific knowledge that must have resulted from indirect evidence?
n200080 [17]

Answer:

The atomic model, the structure of the solar system, and the theory of black holes are examples of models created using indirect evidence.

Explanation:

I have no explanation sorry.

5 0
2 years ago
A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p
stepladder [879]

Answer:

Explanation:

Given that,

Spring constant = 16N/m

Extension of spring

x = 8cm = 0.08m

Mass

m = 5g =5/1000 = 0.005 kg

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

So, Using conservation of energy

Energy in spring is converted to kinectic energy

So, Ux = K.E

Ux = ½ kx²

Then,

Ux = ½ × 16 × 0.08m²

Ux = 0.64 J

Since, K.E = Ux

K.E = 0.64 J

4 0
2 years ago
If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th
cluponka [151]

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

7 0
2 years ago
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i
andrey2020 [161]

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field U=4\times 10^{-6}J

Value of inductance L =0.060 H

Energy stored in magnetic field is given by U=\frac{1}{2}Li^2

i=\sqrt{\frac{2U}{L}}

i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A

So the current flowing through rectangular toroid will be 0.01154 A

3 0
2 years ago
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