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sp2606 [1]
3 years ago
7

A thin, 75.0 cm wire has a mass of 16.5 g. One end is tied to a nail, and the other is attached to a screw that can be adjusted

to vary the tension in the wire. (a) To what tension in Newtons must you adjust the screw to that a transverse wave of wavelength 3.33 cm makes 875 vibrations per second? (b) How fast would this wave travel?
Physics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

(a) 29.14 m/s

(b) 18.7 N

Explanation:

given information:

L = 75 cm = 0.75 m

mass, m = 16.5 g = 1.65 x 10⁻² kg

wavelength, λ = 3.33 cm = 0.0333 m

frequency, f = 875

first we calculate the speed

v = λf

  = 0.0333 x 875

  = 29.14 m/s

v = √F/μ

where

v = speed

F = tension

μ = linear density

μ = m/L, m is mass, and L is the length

thus,

v² = F/(m/L)

F = v²m/L

  = (29.14)²(1.65 x 10⁻² )/(0.75)

  = 18.7 N

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.
GrogVix [38]

Answer:

H = 532 m

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

d = vt

d = 340 \times 3

d = 1020 m

now in the same time the distance that teacher will fall down is given as

d_1 = \frac{1}{2}gt^2

d_1 = \frac{1}{2}(9.81)(3^2)

d_1 = 44.1 m

now total distance traveled by teacher and sound in 3 s

d_{total} = d + d_1

d_{total} = 1020 + 44.1

this total distance must be equal to twice the height of the cliff

2H = 1064.1 m

H = 532 m

7 0
3 years ago
Read 2 more answers
In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

6 0
3 years ago
The time T in seconds for a pendulum of length L feet to make one swing is given by Upper T=2\pi \sqrt((L)/(36)). How long is a
Andreyy89

Answer:

3.6ft

Explanation:

Using= 2*π*sqrt(L/32)

To solve for L, first move 2*n over:

T/(2*π) = sqrt(L/32)

Next,eliminate the square root by squaring both sides

(T/(2*π))2 = L/32

or

T2/(4π2) = L/32

Lastly, multiply both sides by 32 to yield:

32T2/(4π2) = L

and simplify:

8T²/π²= L

Hence, L(T) = 8T²/π²

But T = 2.1

Pi= 3.14

8(2.1)²/3.14²

35.28/9.85

= 3.6feet

5 0
3 years ago
What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes
mr Goodwill [35]
The current is defined as the amount of charge Q that passes through a given point of a wire in a time \Delta t:
I= \frac{Q}{\Delta t}
Since I=500 A and the time interval is
\Delta t=4.0 min=240 s
the charge is
Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C

One electron has a charge of q=1.6 \cdot 10^{-19}C, therefore the number of electrons that pass a point in the wire during 4 minutes is
N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23} electrons
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3 years ago
Q011) The Doppler effect a. occurs when the frequency of sound waves received is lower if the wave source is moving toward you t
inn [45]

Answer:

Option (c) is correct.

Explanation:

The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.

When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.

It occurs for both light and sound.  

So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.

8 0
3 years ago
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