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3 years ago

A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight

Physics

1 answer:

tankabanditka [31]3 years ago

3 0

Answer:

There is no work done.

Explanation:

Given the following data;

Mass = 10 kg

To find the work done?

In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.

Mathematically, work done is given by the formula;

Work done = force * distance

$W = F * d$

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.

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By counting the number of waves that pass a certain point each second, we will know the _______ of the wave.

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The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e

Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ( $v$ ), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

$v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi$ (1)

Where:

$\Delta t$ - Rotation time, measured in seconds.

$R$ - Radius of the Earth, measured in meters.

$\phi$ - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that $\Delta t = 86160\,s$ , $R = 6.371\times 10^{6}\,m$ and $\phi = 37.6819^{\circ}$ , then the tangential speed at Livermore is:

$v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}$

$v\approx 284.001\,\frac{m}{s}$

The tangential speed at Livermore is approximately 284.001 meters per second.

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3 years ago

The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t

hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

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$f=\dfrac{1}{45.4}$

f = 0.022 m

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

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So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

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3 years ago

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj

KATRIN_1 [288]

Answer:

$\Delta t = 8 s$

Explanation:

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so we can use kinematics

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$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

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2 years ago

A train pulls the boxcars. What are the vertical forces present? (1 point)

Greeley [361]

The vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

The forces exerted on an object can be resolved into horizontal and vertical components.

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The frictional force on the object; which tends to reduce the motion,

force dragging the object forward

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This horizontal force is written as;

$\Sigma F_x = 0\\\\F_x - F_k = ma$

The vertical component of the force on the object include the following;

the weight of the object acting downwards

the pull of air on the object acting upwards

The vertical force on the object is written as;

$\Sigma F_n = 0\\\\F_n - W = 0$

Thus, the vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

Learn more here: brainly.com/question/2000189

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2 years ago