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Brut [27]
3 years ago
7

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
1 answer:
Fudgin [204]3 years ago
8 0

Answer:

7.54 m/s

Explanation:

C = 2πr

C = 2π (0.450 m)

C = 2.827 m

v = d / t

v = 25.0 (2.827 m) / (9.37 s)

v = 7.54 m/s

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Absorb kinetic energy and soften potentially damaging blows.
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The release of energy by the nucleus of an atom as a result of nuclear fission is radiant energy
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Energy released by nuclear fission or fusion.
So I think the answer is true.
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3 years ago
A 25-foot ladder is leaning against a house. The base of the ladder is pulled away from the house at a rate of 2 feet per second
grandymaker [24]

A ladder 25 feet long is leaning against a house.  The base of the ladder is pulled away at a rate of 2 ft/sec.  

a.)  How fast is the top of the ladder moving down the wall when the base of the ladder is 12 feet from the wall?

Answer:

dy/dt = -1.094ft/sec

Explanation:

Given that:

dz/dt = 0,

dx/dt = 2,

dy/dt = ?

Hence, we have the following

Using Pythagoras theorem

We have 25ft as the hypotenuse, y as the opposite or height of wall, and x as the base of the triangle

X² + y² = z²,

12² + y² = 25²,

y² = 25² - 12²

y = √481

Therefore, we have the following:

2x dx/dt + 2y dy/dt,

= 2z dz/dt,

= 12 (2) √481 dy/dt,

= √481 dy/dt = -24,

= dy/dt = -1.094ft/sec

Therefore, final answer is -1.094ft/sec

4 0
3 years ago
Grade 11 Physics
xeze [42]
1. Displacement = 50 m North. Distance = 150m
Displacement is a vector. Distance is a scalar.

2. Displacement is 100 miles East.
6 0
4 years ago
A particle with a mass of 9.00 ✕ 10-20 kg is vibrating with simple harmonic motion with a period of 3.00 ✕ 10-5 s and a maximum
Goshia [24]

Answer:

(a) \omega=2.09*10^{5}\frac{rad}{s}

(b) A_{max}=0.033m

Explanation:

(a) The angular frequency is defined as:

\omega=2\pi f

Here f is the frequency of the particle, which is inversely proportional to its period:

f=\frac{1}{T}

Replacing, we have:

\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{3*10^{-5}s}\\\omega=2.09*10^{5}\frac{rad}{s}

(b) The maximum displacement is given by:

A_{max}=\frac{v_{max}}{\omega}\\A_{max}=\frac{7*10^3\frac{m}{s}}{2.09*10^5\frac{rad}{s}}\\A_{max}=0.033m

5 0
3 years ago
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