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sergey [27]
2 years 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]2 years 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]2 years 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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A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum
V125BC [204]
Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)
7 0
2 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

4 0
2 years ago
(ANSWER NOW PLS) A diver with a mass of 90kg is at a height of 10m, and he has not jumped off of the board yet (v=0m/s) what's h
Korolek [52]

Answer: there is zero kinetic energy but there is Gravitational Potential Energy (GPE) and GPE = 8826.3 J

Explanation:

3 0
2 years ago
Manipulate the equation "v=d/t" to find the answers to these problems using
OLEGan [10]
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4 0
3 years ago
Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m
frutty [35]

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

rA = 6.00 m

mB = 62.4 Kg

vB = 4.23 m/s

rB = 3.00 m

mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L =  (mv x r)

⇒  LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒  LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒  LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k  Kgm/s

L = - 1361.591 k Kgm/s

7 0
2 years ago
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