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OLEGan [10]
2 years ago
10

A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a

circular arc rising 0.1m vertically. What must the speed have been just after the collision
Physics
1 answer:
zloy xaker [14]2 years ago
6 0

Answer:

v = 1.4  m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block   is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5  x v² =  9.8 x .1

=  v² = 1.96

v = 1.4

v = 1.4  m /s

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Elena-2011 [213]

Answer:

The minimum speed is 14.53 m/s.

Explanation:

Given that,

r = 11 m

Friction coefficient = 0.51

Suppose we need to find the minimum speed, that the cylinder must make a person move at to ensure they will stick to the wall

When frictional force becomes equal to or greater than the weight of person

Then, he sticks to the wall

We need to calculate the minimum speed

Using formula for speed

f_{s}=\mu N\geq mg

Where, N =\dfrac{mv^2}{r}

\mu\times\dfrac{mv^2}{r}\gep mg

v^2\geq\dfrac{gr}{\mu}

Put the value into the formula

v=\sqrt{\dfrac{9.8\times11}{0.51}}

v=14.53\ m/s

Hence, The minimum speed is 14.53 m/s.

8 0
3 years ago
A trombone plays a C3 note. If the speed of sound in air is 343 m/s and the wavelength of this note is
Umnica [9.8K]

The frequency of note C3 is 131 s^{-1}.

<u>Explanation:</u>

Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.

And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.

Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:

Frequency=\frac{speed of sound}{Wavelength of note}

Thus,

Frequency = \frac{343 m/s}{2.62 m} = 131 s^{-1}

So the frequency of note C3 is 131 s^{-1}.

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Answer:

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You were only given 3 significant figures in the question.

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The answer is d. functionalist perspective 
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