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gayaneshka [121]
3 years ago
15

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ng the axis (middle), with their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s, and an 83kg person feels a 560N force pressing against his back. What is the radius of a chamber
Physics
1 answer:
lana [24]3 years ago
5 0

To solve this problem we will apply the concepts related to centripetal acceleration and Newton's second law. In the case of acceleration, we will define how it is shaped, and we will equate that acceleration to that stipulated by Newton in his second law. We will clear the variable of the 'radius'. Centripetal acceleration is described as,

a_R = \frac{v^2}{r}

Here,

v =Velocity

r = Radius

Now Newton's second law is,

F = ma_R

Here,

m = mass

a_R = Centripetal acceleration

Rearranging in function of the acceleration,

a_R = \frac{F}{m}

Rearranging the first equation in function of the radius

a_R = \frac{v^2}{r} \rightarrow r=\frac{v^2}{a_R}

Equating,

r = \frac{mv^2}{F}

Replacing,

r = \frac{(83kg)(3.2m/s^2)}{560N}

r = 1.518m

Therefore the radius of the  chamber is 1.518m

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4 0
3 years ago
Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N
nadezda [96]
First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say F_{1}, and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
F_{1} = 7i, where i = Unit vector in the x-direction.

2) Take the second force, say F_{2}, and assume that it is making an angle \alpha with the first force F_{1}.

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

F_{2} = (15*cos \alpha)i + (15*sin \alpha )j

Now from simple vector addition, we know that,
F_{R} = F_{1} + F_{2} --- (A)

Where F_{R} = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find F_{1}+F_{2} first!
F_{1}+F_{2} =  7i+(15*cos \alpha)i + (15*sin \alpha )j

=> F_{1}+F_{2} =  (7+15*cos \alpha)i + (15*sin \alpha )j

Now the magnitude of F_{1}+F_{2} is,
| F_{1}+F_{2}| = \sqrt{ (7+ 15*cos \alpha)^{2} +  (15*sin \alpha )^{2}}

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}

Since (sin \alpha)^{2} + (cos \alpha)^{2} = 1, therefore,

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Since  | F_{1}+F_{2}| = |F_{R}|, and the magnitude of the resultant force is 20N, therefore,

 |F_{R}| = | F_{1}+F_{2}|
20 = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Take square on both sides,
400 = 49 + 225 + 210*(cos \alpha)
(cos \alpha) =  \frac{3}{5}

\alpha = 53.13^{o}

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

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Answer:

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