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Cerrena [4.2K]
3 years ago
14

A ball with a weight of 2 N is attached to the end of a cord of length 2m . The ball is whirl in a vertical circle counterclockw

ise. The tension in the cord at the top of the circle is 7N, and at the bottom it is 15 N. (The speed of the ball is not the same in all these points).
a) Three students discuss the net forces on the ball at the top:
Deante: The net force on the ball at the top position is 7 N since the net force is the same as the tension.
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
Fiona: No, you are both wrong. You need to figure out the centripetal force (mv^2/r) and include it in the net force.

Which, if any, of these students do you agree with?

Deante: _____
Edgar: _____
Fiona: _____
None of them: _____
Physics
1 answer:
ZanzabumX [31]3 years ago
8 0

Answer:

we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

Explanation:

Weight of the ball is given as

W = 2N

so we have

m = \frac{W}{g}

m = 0.204 kg

now tension force at the top is given as

T_{top} = 7 N

T_{bottom} = 15 N

Now at the top position by force equation we can say that ball will have two downwards forces

1) Tension force

2) Weight of the ball

so net force on the ball is given as

F_{net} = T + W

F_{net} = 7 + 2 = 9 N

So we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

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3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

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7 0
3 years ago
A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced sli
goblinko [34]

Given Information:  

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

Distance between bright fringes due to laser pointer = yp = 5.14 mm

Required Information:  

Wavelength of the laser pointer = λp = ?

Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d  

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp =  (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

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3 years ago
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Answer:

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6 0
2 years ago
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