Question

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force acting on the object?

a
150 N
b
15 N
c
0 N
d
1.5 N
Question 2 (1 point)
A 3 kg object is moving in circular motion at a speed of 4 m/s. The radius of the circle is 4 m. What is the centripetal force acting on the object?

a
0 N
b
24 N
c
12 N
d
48 N
Question 3 (1 point)
An object is moving in circular motion. The centripetal acceleration of the object is 10 m/s^2. If the radius of the circle is 10 m, what is the speed, v, of the object?

a
0 m/s
b
10 m/s
c
5 m/s
d
100 m/s
Question 4 (1 point)
A race car is driving around a circular track that has a radius of 10 m. If the speed of the car is 5 m/s, what is the centripetal acceleration of the car?

a
2.5 m/s^2
b
5 m/s^2
c
10 m/s^2
d
0 m/s^2
Question 5 (1 point)
An object is moving clockwise in circular motion, as shown in the picture below. When the object is at the position shown in the picture, the centripetal force is...



a
an arrow pointing left.
b
an arrow pointing upward.
c
an arrow pointing downward.
d
an arrow pointing right.
Question 6 (1 point)
A ball is spun clockwise in a circular path, as shown below. When the ball is at the position in the picture, the net force acting on the ball...



a
is an arrow pointing up.
b
is an arrow pointing down.
c
is an arrow pointing left.
d
is 0.
Question 7 (1 point)
A ball is spun counter-clockwise, as shown in the picture below. When the ball is at the position in the picture, the velocity is...



a
0
b
an arrow pointing left.
c
an arrow pointing downward.
d
an arrow pointing upward.
Question 8 (1 point)
If the speed of an object in circular motion is tripled, what change will occur in the centripetal force?

a
The centripetal force will be 9X greater than it was.
b
The centripetal force will be the same.
c
The centripetal force will be 3X greater.
d
The centripetal force will be 1/9 as much as it was.
Question 9 (1 point)
If the speed of an object in circular motion is doubled, what change occurs in the centripetal force?

a
The centripetal force is the same.
b
The centripetal force is 2X bigger.
c
The centripetal force is 4X bigger.
d
The centripetal force is 4X smaller.
Question 10 (1 point)
If the radius of an object in circular motion is doubled, what change will occur in the centripetal force?

a
The centripetal force will be 4X bigger.
b
The centripetal force will be the same.
c
The centripetal force will be 1/2 as big as it was.
d
The centripetal force will be 2X bigger.

Answer 1

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Step-by-step explanation:

1) m = 15kg, a =
`10ms^(-2)`, F = ma = 15*10 = 150N

2) m = 3kg, v =
`4ms^(-1)`, r = 4m, F =
`(mv^(2) )/(r)`

`(3*4^(2) )/(4)` = 12N

3) a =
`10ms^(-2)`, r = 10m, v=?

F =
`(mv^(2) )/(r)` and F = ma

equating the two equations and cancelling a, we have:

`(v^(2) )/(r)` = a

making v the subject of formula, we have:

v =
`√(ar)`

=
`√(100)`

= 10
`ms^(-1)`

4) r = 10m, v =
`5ms^(-1)`, a = ?

F =
`(mv^(2) )/(r)`

F = ma

equating the above equations and making a subject of formula, we get:

a =
`(v^(2) )/(r)`

a = 25/10 = 2.5
`ms^(-2)`

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F =
`(mv^(2) )/(r)`

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F =
`v^(2)`

Now, if v is tripled

F =
`(3v)^(2)`

F = 9
`v^(2)`

We can see that the force will be 9X greater than it was.

9) F =
`(mv^(2) )/(r)`

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F =
`v^(2)`

Now, if v is doubled

F =
`(2v)^(2)`

F = 4
`v^(2)`

We can see that the force will be 4X greater than it was.

10) F =
`(mv^(2) )/(r)`

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was