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Vedmedyk [2.9K]
3 years ago
10

You have a lens whose focal length is 6.91 cm. You place an object on the axis of the lens at a distance of 11.1 cm from it. How

far is the object\'s image from the lens?
Physics
1 answer:
stellarik [79]3 years ago
7 0

Answer:

4.2591 cm

Explanation:

We have given focal length of the lens f=6.91 cm

Object distance u =11.1 cm

We have to find the image distance that is v

For the lens we know that \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

So \frac{1}{6.91}=\frac{1}{v}-\frac{1}{11.1}

0.1447=\frac{1}{v}-0.09

\frac{1}{v}=0.2347

v=4.2591cm

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A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
3 years ago
Which of the following does NOT represent Newton’s second law? Question 20 options: a = m/Fnet m = Fnet/a Fnet = ma a = Fnet/m
Natali [406]

Answer:

a=m/f is not an equation under newton's second law

Explanation:

newton's second law of motion is represented using: f=ma

where a=v-u/t

therefore it becomes,f=m(v-u)/t

from f=ma,

a will become f/m,

m will become f/a

8 0
3 years ago
If the frequency of a given wave increases,what happens to the wavelength?
Greeley [361]
It shortens so that the tips reach faster
8 0
3 years ago
Read 2 more answers
2. A stone is thrown horizontally at a speed of 6.0 m/s from the top of a cliff 78.4 m high.
Ostrovityanka [42]

Answer:

a) 8 seconds if you are using earth's gravity.

b) 48m if the velocity does not change

c) 9.8m/s

Explanation:

3 0
3 years ago
A listener increases his distance from a sound source by a factor of 4.49.
noname [10]

Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

6 0
3 years ago
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