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Romashka [77]
3 years ago
13

an object 8.25 cm from a lens creates a virtual image of magnification 2.40 what is the focal length of the lens (mind your minu

s sign) (unit=cm)
Physics
1 answer:
murzikaleks [220]3 years ago
5 0

Answer:

Focal length of the lens is 14.14 cm

Explanation:

As we know that the image formed due to lens is virtual image

so here we have

M = 2.40

now we have

\frac{d_i}{d_o} = 2.40

now we have

distance of object is 8.25 cm

so we have

d_i = 8.25(2.40) = 19.8 cm

now by lens formula

\frac{1}{d_i} - \frac{1}{d_o} = \frac{1}{f}

\frac{1}{-19.8} - \frac{1}{-8.25} = \frac{1}{f}

f = 14.14 cm

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A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to
daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s                                                                                

Explanation:

It is given that,

Mass of cart 1, m_1=2\ kg

Mass of cart 2, m_2=4\ kg  

Initial speed of cart 1, u_1=3\ m/s          

Initial speed of cart 2, u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}        

V=\dfrac{2\times 3}{(2+4)}

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

J=F\times t=m(V-u)

J=m(V-u)

J=2\times (1-3)    

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0
2 years ago
Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going
DerKrebs [107]

Answer:

The impact force will be same for both the cases.

Explanation:

The rate of change of momentum is known as the Impulse and is given by:

I = \frac{\Delta p}{\Delta t}

where

I = Impulse

\Delta p = change in momentum

\Delta t = time interval

Now,

In first case both the cars are identical and have same velocity and in the second case, the wall is stationary.

Also, in both the cases the car does not bounces off the things it hit.

Thus

\Delta p = 0 - m\times v = - mv

Thus

Impact force, F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}

Therefore, impact force is same for both the cases.

5 0
2 years ago
Read 2 more answers
Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha
MArishka [77]

Answer:

P = 1.64 \times 10^4 Watt

Explanation:

Here we know that the glider is accelerated uniformly from rest to final speed of 25.7 m/s in total distance of d = 46.9 m

so we will have

v_f = 25.7 m/s

v_i = 0

d = 46.9

so for uniformly accelerated motion we have

d = \frac{v_f + v_i}{2} t

46.9 = \frac{25.7 + 0}{2}t

t = 3.65 s

now we will find the total work done given as change in kinetic energy

W = \frac{1}{2}mv^2

W = \frac{1}{2}(181)(25.7^2)

W = 5.97 \times 10^4 J

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P = \frac{W}{t}

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6 0
2 years ago
Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α. express your answer in terms of ω1
VMariaS [17]
T1 = ω1/α.............
3 0
2 years ago
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A gas station owner suspects that he is being overcharged for gasoline deliveries by a gasoline supplier. The overcharge seems p
Tems11 [23]

Answer:

Explanation:

delta V = v * alpha * delta T

= V * 0.00053 * (92.2 - 55.0)

= 0.019716 V

percentage that the owner

= [delta V / V] * 100

= [0.019716 V / V] * 100

= 1.9716 %

4 0
2 years ago
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