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Alex
2 years ago
5

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

Physics
1 answer:
nevsk [136]2 years ago
3 0

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

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

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

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American Football Field Uses A Field That Is 100.0 Yd Long, Whereas A Soccer Field Is 100.0m Long. Which Field Is Longer And By
postnew [5]
Note that
1 yd = 0.9144 m

Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m

Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) =  9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%

Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%
4 0
2 years ago
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
2 years ago
Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of
Tresset [83]

Answer: Well they could go down a hill to gain more kinetic energy, or the answer can just be B. He can pedal harder to increase the rate to 10 meters/second. I hope I helped you.

8 0
1 year ago
Read 2 more answers
A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?
KATRIN_1 [288]

To solve this problem, we use the formula:

I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

 

Therefore the change in intensity from 1m to 100m in decibels is:

B100 – B1 = 10 log(10^-4) dB = -40 dB

 

So the intensity at 100m is calculated as:

B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB

 

Answer:

100 dB

6 0
1 year ago
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You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque
harina [27]

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

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As the axis of rotation is located at the top where they jump, there is a torque.

What is it

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       a = α r

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In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0
1 year ago
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