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A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

y (x, t) = A*cos(k*x -w*t)

Where k = stiffness and w = angular frequency

Hence,

k = 172 and w = 2730

- Calculate wave speed V:

V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

T = W = Y*V^2 = (w/L*g)*V^2

W = (0.0126 / 1.8*9.81)*(13.78)^2

W = 0.135 N

4 0

4 years ago

The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find

snow_tiger [21]

Answer:

The emf of the generator is 6V

The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7 + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7 + 0.3r

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0

3 years ago

You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)

Jet001 [13]

Answer:

1) the correct answer is b and d

2) For force b its direction is vertical up

for the force d its direction is vertical down

3) the correct answers are: a, c and d

4) Force a is vertical down , force c is vertical up and force d is vertical down

5) the correct answer which is b

Explanation:

In this exercise it is asked to identify the forces, fundamentally on the free there are the forces of gravity and the support force of the hand, with these facts we answer the questions

1) the correct answer is b and d

the hand acts on the book with a contact force and the Earth acts on the book with the force of gravity.

2) For force b its direction is vertical up

for the force d its direction is vertical down

3) The forces on the hand are the weight of the book. The force of gravity due to the mass of the hand. As the hand is in balance, there must be a force applied by the arm to keep the hand in position; assuming the hand is in the air, if the hand is resting on the floor the force of the floor on the hand can perform this function

therefore the correct answers are: a, c and d

4) Force a is vertical down

force c is vertical up

force d is vertical down

5) The action and reaction forces are forces of equal magnitude, each applied to one of the bodies, we have

* the force of the hand on the free and its reaction the force of the book on the hand

* The force of the Earth on the book and the hand, giving the weight of each one and the relationship is the force of the book and the hand on the Earth

the correct answer which is b

8 0

3 years ago

A 65kg person throws a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the

asambeis [7]

Answer:

v₁ = 2.48m/s, v₂ = 0.02m/s

Explanation:

Momentum p must be conserved. p = mv

1) First person throwing the snow ball. The momentum before the throw:

p = (65kg + 0.045kg) * 2.5 m/s

The momentum after the throw:

p = 65kg * v₁ + 0.045kg * 30m/s

Solving for the velocity v₁ of person 1:

v₁ = ((65kg + 0.045kg) * 2.5 m/s - 0.045kg * 30m/s) / 65kg = 2.48m/s

2) Second person catching the ball. The momentum before the catch:

p = 0.045kg * 30m/s + 60kg * 0m/s

The momentum after the catch:

p = (60kg + 0.045kg) * v₂

Solving for velocity v₂ of person 2:

v₂ = 0.045kg * 30m/s / (60kg + 0.045kg) = 0.02 m/s

5 0

3 years ago

Which example shows potential energy? A. a skydiver falling B. a car racing C. hitting a nail with a hammer D. a wound up watch

tankabanditka [31]

B.a car racing down a hill

6 0

3 years ago

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