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kherson [118]
3 years ago
6

-g With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to

have a wavelength of 0.750 m?
Physics
1 answer:
Alex17521 [72]3 years ago
8 0

Answer:

43.2 N

Explanation:

\lambda = Wavelength = 0.75 m

f = Frequency = 40 Hz

m = Mass of string = 0.12 kg

L = Length of string = 2.5 m

\mu = Linear density = \dfrac{m}{L}

Velocity of wave is given by

v=\lambda f\\\Rightarrow v=0.75\times 40\\\Rightarrow v=30\ m/s

The tension in string is given by

T=v^2\mu\\\Rightarrow T=30^2\times \dfrac{0.12}{2.5}\\\Rightarrow T=43.2\ N

The tension in the string is 43.2 N

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A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c
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(a) +9.30 kg m/s

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I= \Delta p = m \Delta v = m (v-u)

where

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\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

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parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet
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Answer:

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

- Te energy density u = 0.037 J /m^3

Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

                               u = 0.5*ε*E^2

- Where the Electric field strength E between capacitor plates is given by:

                               E = V / d

Hence,

                               u = 0.5*ε*(V/d)^2

Where, ε = 8.854 * 10^-12

                               V^2 = 2*u*d^2 / ε

                               V = d*sqrt ( 2*u / ε )

Plug in values:

                               V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )

                               V = 576 V

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