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MrRissso [65]
3 years ago
11

A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?

Physics
2 answers:
Elis [28]3 years ago
5 0

Answer: 7.544 m/s

Explanation: Go search it on yahoo if you need an explanation.

Sauron [17]3 years ago
3 0

The velocity of the ball is 12.5 m/s

Explanation:

The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:

v=\frac{d}{t}

First, we calculate the distance covered. We know that the radius of the circle is

r = 0.450 m

And the length of the circumference is

L=2\pi r = 2\pi(0.750)=4.7 m

The ball makes 25.0 revolutions, so a total distance of

d=(25.0)L=(25.0)(4.7)=117.5 m

In a time of

t = 9.37 s

So, its velocity is

v=\frac{117.5}{9.37}=12.5 m/s

Learn more about velocity here:

brainly.com/question/5248528

#LearnwithBrainly

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Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi
Lemur [1.5K]

Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

MaVa+MbVb = MaVa+MbVb (after collision)

Substitute the given values

5(3)+10(0) = 5Va+10(2)

15+0 = 5Va + 20

5Va = 15-20

5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

6 0
3 years ago
How many newtons does a 5-kg backpack weigh on Earth?
zvonat [6]
A 5kg backpack will weigh 49 newtons on earth
7 0
3 years ago
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers
jek_recluse [69]

Answer:

0.74 m/s

Explanation:

From the question,

We apply the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Since the skateboard, the person and the brick where stationary, therefore, the total momentum before collision is 0

0 = Total momentum after collision

(m+M)V + m'v = 0

Where m = mass of the  skateboard, M = mass of the person, m' = mass of the brick, V = recoil velocity of the person and the skateboard, v =  velocity of the brick

make V the subject of the equation above

V = -m'v/(m+M)................... Equation 1

Given: m = 4.10 kg, M = 68.0 kg, m' = 2.50 kg, v = 21.0 m/s.

Substitute these values into equation 1

V = -(2.5×21)/(68+2.5)

V = 52.50/70.5

V = 0.74 m/s

4 0
3 years ago
Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli
Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

Explanation:

Given:

Mass of  Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = v_{f}=?

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}

Substituting the values we get

2\times 14 + 5\times 0 =(2+5)\times v_{f}

v_{f}=\dfrac{28}{7}=4\ m/s

Therefore,

Final velocity of the coupled carts after the collision is

v_{f}=4\ m/s

8 0
3 years ago
The balmer series is formed by electron transitions in hydrogen that
RSB [31]

Answer:

The Balmer series refers to the spectral lines of hydrogen, associated to the emission of photons when an electron in the hydrogen atom jumps from a level n \geq 3 to the level n=2.

The wavelength associated to each spectral line of the Balmer series is given by:

\frac{1}{\lambda}=R_H (\frac{1}{2^2}-\frac{1}{n^2})

where R_H is the Rydberg constant for hydrogen, and where n is the initial level of the electron that jumps to the level n = 2.

The first few spectral lines associated to this series are withing the visible part of the electromagnetic spectrum, and their wavelengths are:

656 nm (red, corresponding to the transition 3 \rightarrow 2)

486 nm (green, 4 \rightarrow 2)

434 nm (blue, 5 \rightarrow 2)

410 nm (violet, 6 \rightarrow 2)

All the following lines lie in the ultraviolet part of the spectrum. The limit of the Balmer series, corresponding to the transition \infty \rightarrow 2, is at 364.6 nm.

4 0
4 years ago
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