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lidiya [134]
3 years ago
8

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

he fundamental frequency on a piece of the same string that is twice as long and has four times the tension
Physics
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}    .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}    .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

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ArbitrLikvidat [17]

Gas always expands or contracts to exactly fill whatever you put it in.

So to measure the volume of a gas, just measure the volume of the jar, the tank, the bottle, the can, or the balloon that the gas is in.

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3 years ago
A concave mirror with a radius of curvature of 20 cm has a focal length of
xxTIMURxx [149]

Answer:

A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

The focal length is of 10 cm, object distance is 30 cm and magnification is -2.

Explanation:

Given:

A concave mirror:

Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

a.

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror =  \frac{C}{2} = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

b.

We have to find the object distance:

Formula to be use:

⇒ \frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}

⇒ Plugging the values.

⇒ \frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}

⇒ \frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}

⇒ \frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}

⇒ \frac{-3+2}{30} =\frac{1}{image\ distance}

⇒ \frac{-1}{30} =\frac{1}{image\ distance}

⇒ -30\ cm=image\ distance

Image will be formed towards negative x-axis 30 cm away from the pole.

c.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ m=-\frac{image\ distance}{object\ distance}

⇒ m=-\frac{(-30)}{(-15)}

⇒ m=-2

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

5 0
3 years ago
student built a simple heat engine that could lift masses. If the heat engine takes in 300 J of heat and loses 50 J to the envir
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Answer:

So it will lift the mass by h = 17 m

Explanation:

As per energy conservation we know that

Q_1 = W + Q_2

here we know that

Q_1 = 300 J

Q_2 = 50 J

now we have

W = 300 - 50

W = 250 J

so work done by the engine is 250 J

now we have

W = mgh

250 = 1.5 \times 9.81 \times h

h = 17 m

4 0
3 years ago
A 3kg object moving at 4 m/s encounters a 20 N resistive force over a duration of 0.20 seconds. The impulse experienced by this
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Answer:

the impulse experienced by this object is 4 Ns

Explanation:

Given;

mass of the object, m = 3 kg

velocity of the object, v = 4 m/s

resistive force, F = 20 N

duration of impact, t = 0.2 s

The impulse experienced by this object is calculated as follows;

J = F x t

J = 20 x 0.2

J = 4 Ns

Therefore, the impulse experienced by this object is 4 Ns

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3 years ago
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Leya [2.2K]

A reference is critical because it determines how much motion has occurred. You can't determine how much an objected has moved if you don't know where it came from.

3 0
3 years ago
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