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77julia77 [94]
3 years ago
7

What is the best explanation of work?

Physics
2 answers:
9966 [12]3 years ago
7 0

Answer:

In  physics, work is defined as a  force  causing the movement—or displacement—of an object. In the case of a constant force, work is the scalar product of the force acting on an object and the displacement caused by that force.

Contact [7]3 years ago
7 0

Answer: option D

Explanation:  In physics, you say that a force is "doing work" when the force is acting on an object and causes a movement in the said object. This means that if you start pushing a wall, and the wall does not move, there is no work generated, indifferent of the amount of force applied.

Under this viewpoint, the right option is D.

"Work occurs when a force acts on an object and causes a displacement of that object."

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An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

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

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

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

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
3 years ago
If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for
steposvetlana [31]

Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

5 0
3 years ago
The US Environmental Protection Agency issues a daily report for pollution levels called the __________.
VARVARA [1.3K]
I think it’s D lol i’m not sure
5 0
4 years ago
Read 2 more answers
Distance measurements based on the speed of light; used for objects in space
Vika [28.1K]

Answer : Yes, distance measurements based on the speed of light used for objects in space.

Explanation :  A light year is measurement  of distance  that light travel in a one year.

In a one year light travels 9460000000000 kilometer.

We know that, speed of light is 3\times10^{8}\ m/s

and  time is 31536000 seconds in 1 year

so, distance = speed of light X time

Now, the light year is 9.4608\times10^{15} meter

Example : The nearest star to earth is about 4.3 light year away.

4 0
3 years ago
A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
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