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kipiarov [429]
3 years ago
11

An object is 10. cm from the mirror, its height is 3.0 cm, and the focal length is 2.0 cm. What is the distance from the

Physics
1 answer:
Rashid [163]3 years ago
4 0

The distance from the image to the mirror is C) 2.5 cm

Explanation:

A mirror is an object that reflects the light coming from an object, creating an image of the object itself.

We can solve this problem by using the mirror equation, which states that:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f is the focal length of the mirror

p is the distance between the object and the mirror

q is the distance between the image and the mirror

For the situation in  this problem, we have:

f = 2.0 cm is the focal length of the mirror

p = 10.0 cm is the distance of the object from the mirror

Solving for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{2.0}-\frac{1}{10}=\frac{4}{10}\\\rightarrow q=\frac{10}{4}=2.5 cm

Learn more about reflection and other wave phenomena:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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What is the SI unit of electric charge
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Answer:

The SI unit for electric chargeis the C (which is the abbreviation of Coulomb}

Explanation:

The SI unit for electric chargeis the Coulomb. The letter used is the C.

1 C = 1 As

4 0
3 years ago
An object with a mass of 5kg accelerates at 2m/s2. How much force in Newtons(N) is needed to cause this to happen??
Reptile [31]

Answer: The formula of Newtons second law of motion is F=MA so therefore it would be written like this Force = Mass X Acceleration

F = 5 x 2

F = 10 N

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3 years ago
What happens to rocks as water combines
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The answers is
D. The acid creates cracks in the rocks, which
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3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
3 years ago
A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
Hoochie [10]

The acceleration of the box is approximately 1 m/s^2

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

\sum F = F_a - F_f

where

F_a = 20 N is the applied forward force

F_f=9.0 N is the friction force

Combining the two equations,

F_a-F_f=ma

And solving for the acceleration,

a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2

Learn more about Newton's second law:

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