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hodyreva [135]
3 years ago
9

A wave on a string is described by D(x,t)= (2.00cm)sin[(12.57rad/m)x−(638rad/s)t], where x is in m and t is in s. The linear den

sity of the string is 5.00g/m.
What is the string tension?

What is the maximum displacement of a point on the string?

What is the maximum speed of a point on the string?
Physics
1 answer:
mamaluj [8]3 years ago
5 0

Answer : The string tension is T = 12.882 N

The maximum displacement is 0.02 m

The maximum speed is v = 12.76\ m/s

Explanation :

Given that,

D(x,t) = (2.00 cm) sin [(12.57rad/m)x - (638rad/s)t]

Where, x is in m and t is in sec.

Linear density of the string = 5.00 g/m

We know that,

Velocity of the wave

v = \dfrac{\omega}{k}

v = \dfrac{638}{12.57}\ m/s

v = 50.76\ m/s

Now, the string tension

v = \sqrt\dfrac{T}{m}

T = 0.005\times (50.76)^{2}\ N

T = 12.882 N

The maximum displacement is 0.02 m

The maximum speed  

v = a \omega

v = 0.02\times638\ m/s

v = 12.76\ m/s

Hence, this is the required solution.





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GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h  = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between  ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0
3 years ago
The rotational speed of earth is similar to?​
m_a_m_a [10]
Actually, the speed of the earth is the same everywhere, taking the angular speed as the valid measure of the speed
8 0
3 years ago
A locomotive approaches its next stop and accelerates at -0.12 m/s^2, coming to a complete stop in 30 seconds. This motion could
Masja [62]

Answer:

<em>Answer: positive velocity & negative acceleration</em>

Explanation:

<u>Accelerated Motion</u>

Both the velocity and acceleration are vectors because they have magnitude and direction. When the motion is restricted to one dimension, i.e. left-right or up-down, the direction is marked with the sign according to some preset reference.

The locomotive is moving at a certain speed with a (so far) unknown sign but the acceleration has a negative sign. Since the locomotive comes to a complete stop it means the velocity and the acceleration are of opposite signs.

Thus the velocity is positive.

Answer: positive velocity & negative acceleration

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He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How
Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

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velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

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The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

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The mass of the space shuttle is calculated as follows;

W = mg\\\\m = \frac{W}{g} \\\\m = \frac{750,000}{9.8} \\\\m = 76,530.61 \ kg

The force needed to keep the space shuttle moving at constant speed is calculated as follows;

F = ma

F = 76,530.61 \times a

where;

a is the acceleration of the space shuttle

At a constant speed, acceleration is zero.

F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

Learn more here:brainly.com/question/16374764

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