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
NH4OH is the answer. Hope this helps you.
Answer:
The heavier piece acquired 2800 J kinetic energy
Explanation:
From the principle of conservation of linear momentum:
0 = M₁v₁ - M₂v₂
M₁v₁ = M₂v₂
let the second piece be the heavier mass, then
M₁v₁ = (2M₁)v₂
v₁ = 2v₂ and v₂ = ¹/₂ v₁
From the principle of conservation of kinetic energy:
¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J
¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400
¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400
K.E₁ + ¹/₂K.E₁ = 8400
Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁
1.5 K.E₁ = 8400
K.E₁ = 8400/1.5
K.E₁ = 5600 J
K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J
Therefore, the heavier piece acquired 2800 J kinetic energy
Answer:
It does both. Once they get close enough the air does start to get charged, but then they eventually discharge when they touch.
Explanation:
I would say that I and IV because summer is I and fall is IV
