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Dahasolnce [82]

1 year ago

8

A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

ork and the block slides a distance s along the incline before it stops. Determine the value of s.

1 answer:

Musya8 [376]1 year ago

5 0

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

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A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if

saul85 [17]

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

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1 year ago

Which of the following statements is false?

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A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.

Virtual images are property of concave lenses. These images appear closer but smaller than the real object.

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2 years ago

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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

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1 year ago

The image shows a pendulum that is released from rest at point A. Shari tells her friend that no energy transformation occurs as

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1 year ago

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A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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2 years ago