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mariarad [96]
3 years ago
7

Could someone help me with these two problems please?! Will give brainiest (if possible)!

Physics
1 answer:
lubasha [3.4K]3 years ago
3 0
A screw is c) inclined plane

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A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be
Troyanec [42]
Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement 
so when we wanted to move a 100kg a distance of 1m 
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m 

by the same way:-
------------------------
work in = 100kg * 2m * g (m/s^2)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
6 0
3 years ago
Read 2 more answers
The low-_____ areas in a sound wave are called rarefactions. A. period B. frequency C. energy D. density
Delicious77 [7]
You can picture a sound wave a lot like a Slinky wave . . . a
thicker, compressed blob moving along the path, with thinner,
stretched-out places before and after it.

The thicker parts of a sound wave, where the air is more dense,
are called compressions.

The thinner parts of a sound wave, where the air is less dense,
are called rarefactions.
4 0
3 years ago
Read 2 more answers
6) What force should be applied to compress a spring
valina [46]

Answer:

26.6N

Explanation:

Using Hooke’s law

F = kx

Where F = force

K = spring constant

x = displacement

From the question

F= ?

K = 140 N/m

x = 0.19m

Therefore,

F = 140 x 0.19

= 26.6N

7 0
3 years ago
Nichrome is used as heating element in heater.Why?​
Masja [62]

Nichrome is a mixture of nickel and chromium. Nichrome Wire is highly resistant to the flow of electricity . This resistance causes an increase in generated heat . It also has the ability to be heated many times without being destroyed. It is is a mixture of nickel and chromium.This heat is powerful enough to heat the water without damaging the wire.

7 0
4 years ago
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