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laila [671]
2 years ago
11

You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H

z , creating a continuous sinusoidal wave of wavelength 1.0 m. Part A How much time will pass until a standing wave fills the entire length of the string
Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
7 0

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end x= 3m, frequency of the wave f = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ v = fλ

Where v = speed of the standing wave.

also, ∴ v=\frac{x}{t}

where t = time take to fill entire length of the string.

Compare above both equation,

⇒   t = \frac{3}{2.3} sec

     t = 1.3sec

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

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A uniformly charged conducting sphere of 1.3 m diameter has a surface charge density of 8.1 µc/m2. (a) find the net charge on th
8_murik_8 [283]
<span>The surface charge density = q/A So q = surface charge density x Area The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998. So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6) The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space. So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)</span>
8 0
2 years ago
magnitude of velocity does not change but the direction changes. Can we say that it is an accelerated motion?
Olenka [21]

Answer:

MRCORRECT has answered the question

Explanation:

Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed ordirection, or both. Keep in mind that althoughacceleration is in the direction of the changein velocity, it is not always in the direction ofmotion.

4 0
3 years ago
An amateur golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for o
Rudiy27

Answer:

The average force ≅ 519.44 N.

Explanation:

Impulse = change in momentum of a body

i.e Ft = m(v - u)

where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.

m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s

So that,

F x 0.00360 = 0.055(34 - 0)

F x 0.00360 = 0.055 x 34

                    = 1.87

F = \frac{1.87}{0.0036}

 = 519.4444

The average force exerted on the ball by the club is approximately 519.44 N.

4 0
3 years ago
Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t
Step2247 [10]

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

change in pressure  for oil is given as

\Delta p = \rho gh

here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

change in pressure  for WATER is given as

\Delta p = \rho gh

here density is for water and h is for water

\Delta p = 1000*5 *9.81 = 49,050 kPa

pressure change due to both is given as

\Delta p = 41692.3 + 49050 = 90742.5 N/m2

\Delta p = 90.7 kPa

8 0
3 years ago
Read 2 more answers
A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the
Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>(Either option A or B without the minus symbol before </em>F_{bat}<em>, likely typographical error)</em>

<em />

Reason:

Force exerted on the bat by the ball = F_{ball}

Force exerted on the ball by the bat = F_{bat}

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, F_{bat}, is reacted to by the force of the ball acting on the bat, F_{ball}

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>Where the force of the bat is high, the ball is accelerated to travel at high speed</em>

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0
2 years ago
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