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3 years ago

Can someone check my answers and tell me if their correct?

Physics

1 answer:

Otrada [13]3 years ago

6 0

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi = 2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

Formula

a = (vf - vi)/t

Solve

-9.81 = (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0 m/s But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

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