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alina1380 [7]
3 years ago
15

A horizontal 2.0 x 101 N forces pushes an object along a level surface at a constant speed of 3.0m/s for 7.0 s. How much work is

done?
a. 21 J b. 60 J c. 140 J d. 420 J
Physics
1 answer:
Leona [35]3 years ago
8 0

Answer:

420J

Explanation:

Given parameters:

Force  = 2 x 10¹N

Speed  = 3m/s

Time  = 7s

Unknown:

Work done  = ?

Solution:

Work done is the force applied on a body to move it through a particular distance.

   Work done  = force x distance

Now, let us find the distance;

    distance  = speed x time

    distance  = 3 x 7 = 21m

So;

    Work done  = 2 x 10¹ x 21  = 420J

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A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2
Rudiy27

Answer:

Approximately 7.66\; \rm m.

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of u = 1.75\; \rm m \cdot s^{-1} and came to a stop when her speed became v = 0\; \rm m \cdot s^{-1}. How much time would that take if her acceleration is a = -0.20\; \rm m \cdot s^{-1}?

\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}.

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m.

In other words, the skateboarder travelled 15.3\; \rm m up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a},

where:

  • u and v are the initial and final velocity of the object,
  • a is the constant acceleration that changed the velocity of this object from u to v, and
  • x is the distance that this object travelled while its velocity changed from u to v.

For the skateboarder in this question:

\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}.

6 0
4 years ago
Which structure controls how much light passes through the specimen
poizon [28]

Answer:

The diaphragm.

Explanation:

A diaphragm is a thin non transparent structure with an aperture at its center. Aperture is the opening in a lens through which light passes to enter the camera. Diaphragm controls the passage of light through specimen. It stops the passage of light except for the light passing through aperture. It also limits the brightness of light reaching the focal plane.

The diaphragm is placed close to the lens, where objects are defocused to the maximum in order to pass every ray from the object through the lens. Diaphragm discards some of those rays but allows multiple rays to move through to produce an image. This means that the size of the aperture controls the amount of light that passes through the lens. The center of the aperture coincides with optical axis of the lens.  Iris diaphragm is an example. It is used in modern cameras.

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