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2 years ago

You are sitting on the beach and wondering about the properties of mechanical waves. Describe them in terms of ocean waves.

Physics

1 answer:

Rudiy272 years ago

7 0

Answer:

The ocean waves is a mechanical wave that transmits mechanical energy in the wave by the synchronized and repeated oscillation of the waters about an equilibrium level such that as the wave approaches the shoreline, and the water depth decreases, the height of the wave also increases reflecting the effective transmission of energy while the medium which is the water through which the wave propagates, move back and forth within a small region

Explanation:

A mechanical wave like other waves is the oscillation of a field about an equilibrium level. In mechanical waves, the field consists of the oscillating matter such that the wave transmits energy through a medium. The displacement of the medium through which the wave energy is limited such that the wave energy is conserved to travel far.

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How much power does it take to lift a 24 kg gift box 6m above the floor in 4 s?

Mrac [35]

Answer:

Explanation:

Step one:

given

mass m= 24kg

distance moved= 6m

time taken= 4seconds

Step two:

Required

power

but work done is the force applied at a distance, and the power is the work done time the time taken

Work done= F*D

F=mg

W= mg*D

W=24*9.81*6

W=1412.6J

Power P= work * time

P=1412.6*4

p=5650.5W

P=5.6kW

3 0

2 years ago

The primary of a step-up transformer is connected across the terminals of a standard wall socket, and resistor 1 with a resistan

alexira [117]

The turns ratio is the factor that determines voltage andcurrent. In order to have the same current across the resistorin the primary as the resistor in the secondary, then:--N(p) = Primary turnsN(s) = Secondary turnsR(2) = Primary resistorR(1) = Secondary resistor--R(2)/R(1) = N(p)/N(s)R(2) = R(1)*(N(p)/N(s))--If arbitrary values are plugged in, you will see that this step up transformer will require 2x the resistance required in the secondary, R(1), to obtain the same current. Thus R(2) will be 1/2 the value of R(1). This is due to the stepped up voltage in the secondary.

3 0

2 years ago

A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of

Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = $20\times 2000\ lb$

d = Diameter = 1.25 in

$L_1$ = Initial length = 50 ft

$L_2$ = Final length

A = Area = $\dfrac{\pi}{4}d^2$

Y = Young's modulus = $30\times 10^6\ psi$

Young's modulus is given by

$Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft$

The length during the lift is 50.0543248872 ft

6 0

2 years ago

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

enyata [817]

Answer:

Emitted power will be equal to $7.85\times 10^{-5}watt$

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing $I=10^{-12}W/m^2$

We have to find the emitted power

Emitted power is equal to $P=I\times A$

$P=I\times 4\pi R^2$

$P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt$

So emitted power will be equal to $7.85\times 10^{-5}watt$

4 0

2 years ago

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

2 years ago