The image distance when a boy holds a toy soldier in front of a concave mirror, with a focal length of 0.45 m. is -0.56 m.
<h3>What is image distance?</h3>
This is the distance between the image formed and the focus when an object is placed in front of a plane mirror.
To calculate the image distance, we use the formula below.
Formula:
- 1/f = 1/u+1/v........... Equation 1
Where:
- f = Focal length of the mirror
- v = Image distance
- u = object distance
From the question,
Given:
Substitute these values into equation 1 and solve for the image distance
- 1/0.45 = 1/0.25 + 1/v
- 2.22 = 4+1/v
- 1/v = 2.22-4
- 1/v = -1.78
- v = 1/(-1.78)
- v = -0.56 m
Hence, The image distance is -0.56 m.
Learn more about image distance here: brainly.com/question/17273444
Answer:
isoceles
Explanation:
Because two sides are equal
The amount of power change if less work is done in more time"then the amount of power will decrease".
<u>Option: B</u>
<u>Explanation:</u>
The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt
Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.
This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but when the work is done in a shorter period of time more power is required.
Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,

For 8 shot burst, average recoil force on the gun is :

So, the average recoil force on the gun during that 0.40 s burst is 45 N.