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3 years ago

A camera employs _lens to form_images

Physics

1 answer:

noname [10]3 years ago

8 0

Answer:

a camera employs camera lens to firm some images.

Explanation:

hope this helps.

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What is work done aganist gravity?

Rudik [331]

Answer:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity.

Explanation:

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3 years ago

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What is the difference between convex and concave mirrors and lens?

Sidana [21]

Concave makes things smaller and convex makes things bigger

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3 years ago

HELP!

Goryan [66]

Answer:

The solved problem is in the photo. Hope it helps.

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3 years ago

Summarize: Based on what you have learned, how will the sound that the observer hears

Reil [10]

Answer: The sound will change due to changes in frequency and the wavelength of the airplane.

Explanation: Let assume that the observer is at a stationary position. The wavelength of the sound from the airplane reduces and the frequency increases as the plane is moving toward the observer. As the airplane passes by, that is, moving away from the observer, the frequency starts to reduce while the wavelength of the sound starts to increase.

The sound that the observer hears will change base on the illustration above.

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3 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

A camera employs _lens to form_images ​

A camera employs _lens to form_images