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igor_vitrenko [27]
3 years ago
11

2. A ball tied to a pole by a rope swings in a circular path with a centripetal acceleration of 2.7 m/s. If the ball has a

Physics
1 answer:
Helga [31]3 years ago
3 0

Answer: The diameter of the circular path is 2.96m

Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.

Centripetal acceleration = 2.7m/s^2

Tangential speed = 2.0m/s

Radius = 2.0^2 / 2.7 = 4/2.7

= 1.48m

Diameter = radius*2

= 1.48*2 = 2.96m.

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Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

T=\dfrac{mv^2}{r}+mg

\dfrac{T}{m}-g=\dfrac{v^2}{r}

v=\sqrt{(\dfrac{T}{m}-g)r}

v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

8 0
3 years ago
How large a band of frequencies does each television broadcasting channel get ?
k0ka [10]

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

8 0
2 years ago
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A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

a'=4\times \dfrac{v^2}{R}

a'=4\times a

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
A 30 kg student drops down from the monkey bars. The acceleration due to gravity is -9.8 m/s^2. Neglecting air drag, what is the
Sholpan [36]

The net force on the student is A) -294 N

Explanation:

Neglecting air resistance, there is only one force acting on the student: the force of gravity, which is given by

F=mg

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m is the mass of the student

g is the acceleration of gravity

In this problem, we have:

m = 30 kg is the mass of the student

g=-9.8 m/s^2 is the acceleration of gravity, where the negative sign means the direction is downward

Substituting, we find the force of gravity on the student:

F=(30)(-9.8)=-294 N

And since this is the only force acting on the student, it is also the net force on him.

Learn more about gravitational force:

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8 0
3 years ago
Light has a frequency of about 6 x1014 Hz.
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Answer:

The visible light frequency is 400 THz to 700 THz, approximately. A THz is a Terahertz, which is a unit of frequency equal to one trillion Hertz.

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2 years ago
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