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3 years ago

Franny drew a diagram to compare images produced by concave and convex lenses.

Physics

2 answers:

Stella [2.4K]3 years ago

7 0

Answer:

B. virtual

Explanation:

Answer on edge 2020

yanalaym [24]3 years ago

4 0

Answer:

virtual

Explanation:

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A rocket passes you at a speed of 0.85c, and you measure its length to be 35.0 m. What is its measured length when at rest?

viktelen [127]

Answer:

66.4 m

Explanation:

To solve the problem, we can use the length contraction formula, which states that the length observed in the reference frame moving with the object (the rocket) is given by

$L=L_0 \sqrt{1-(\frac{v}{c})^2}$

where

$L_0$ is the proper length (the length measured from an observer at rest)

v is the speed of the object (the rocket)

c is the speed of light

Here we know

v = 0.85c

L = 35.0 m

So we can re-arrange the equation to find the length of the rocket at rest:

$L_0 = \frac{L}{\sqrt{1-(\frac{v}{c})^2}}=\frac{35.0}{\sqrt{1-(\frac{0.85c}{c})^2}}=66.4 m$

8 0

4 years ago

A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),

labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

FOR FORCE (P):

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

Therefore, work done by force (P) is Positive.

FOR NORMAL FORCE (Fn) AND WEIGHT (W):

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.

FOR KINETIC FRICTIONAL FORCE (fk):

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

Therefore, work done by Kinetic Frictional Force (fk) is Negative.

8 0

4 years ago

The law of inertia to both moving and no moving objects. True or false

Zepler [3.9K]

True, the law of inertia effects both moving and non-moving objects.

3 0

3 years ago

Read 2 more answers

How long would it take a plane to fly 1,054 km with an average speed of 886 km/h?

Leviafan [203]

Answer:

1.19 hours

Explanation:

divide distance by speed. hope this helps

8 0

2 years ago

A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of

Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0

3 years ago