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3 years ago

5

A fire helicopter carries a 580-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

d of 40.0 m/s, the cable makes an angle of 38.0° with respect to the vertical. Determine the force exerted by air resistance on the bucket.

1 answer:

inn [45]3 years ago

4 0

Answer:

F = 41,954 N

Explanation:

given,

mass of bucket = 580 Kg

length of the cable = 20 m

velocity = 40 m/s

angle made = 38.0°

T cos 38° = m g..............(1)

T sin 38^0 = \dfrac{mv^2}{l} + F......(2)

dividing equation (2) by (1)

$tan 38^0 = \dfrac{\dfrac{mv^2}{l} + F}{mg}$

$tan 38^0 = \dfrac{\dfrac{580\times 40^2}{20} + F}{580 \times 9.81}$

$4445.36 = \dfrac{580\times 40^2}{20} + F$

F = -46400 + 4445.36

F = -41,954 N

hence, the force is acting in the opposite direction as assumed.

F = 41,954 N

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Pls help me… I missed the whole lesson and I have no clue what to do-

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Answer:

1) 50 facing towards the right

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Explanation:

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2 years ago

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

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$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

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2 years ago

if the angular momentum of a rigid body is changing, does that mean that there must be a net torque acting on the body?

masha68 [24]

The attribute of any rotating object determined by the product of the moment of inertia and the angular velocity is known as angular momentum.

<h3>What is Angular Momentum?</h3>

Without a kickstand, attempting to balance while getting on a bicycle will definitely result in you falling off. However, these wheels gain angular momentum once you begin pedaling. They're going to be resistant to change, which will make balance simpler.
The definition of angular momentum is: any rotating object's characteristic determined by moment of inertia times angular velocity.
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As a result, she conserves angular momentum and spins faster.

To Learn more About angular momentum refer to :

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6 0

10 months ago

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

So

$m = 25 * 12.5$

$m \approx 310$

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3 years ago

If a cart of a roller coaster has a mass of 250kg and is at a height of 14 meters. What is the cart's potential energy?

ahrayia [7]

Answer:

3430000 J

Explanation:

The formula for potential energy is PE=mgh.

M being the mass, g being the force of gravity, and h being the height.

First thing you want to do is convert 250 kg to g (grams).

From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).

5 0

3 years ago