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1 year ago

A 6.00-m long rope is under a tension of 600 N. Waves travel along this rope at 40.0 m/s. What is the mass of the rope?

Physics

1 answer:

MAVERICK [17]1 year ago

6 0

Answer:

2.25 kg

Explanation:

Length, L = 6 m

tension, T = 600 N

velocity, v = 40 m/s

The formula for the velocity is given by

$v=\sqrt{\frac{T}{\mu }}$

where, μ is mass per unit length

$\mu =\frac{T}{v^{2}}$

$\mu =\frac{600}{40^{2}}$

μ = 0.375

m / L = 0.375

m = 0.375 x 6 = 2.25 kg

Thus, the mass of rope is 2.25 Kg.

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Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,

VARVARA [1.3K]

Answer: yes.

Explanation: The light that will be incidented on that metal is visible light.

It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer.

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2 years ago

The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

Work done by brake = 1000 × 10 × 100

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

4 0

2 years ago

Vader's light saber is red, while Obi-Wan's light saber is blue, meaning that Obi-Wan's light saber is emitting _____ compared t

Bingel [31]

I assume it woukd be higher energy light waves. when fire is at its hottest state its blue because its burning off so much.

4 0

2 years ago

Read 2 more answers

What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is $W=mg=2.28 N$

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

$F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N$

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

$F_y=W=mg=2.28 N$ (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

7 0

2 years ago

Read 2 more answers

A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the

Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0

1 year ago