Answer:
<h2>Virtual image</h2>
Explanation:
<h3>
<em>Virtual</em><em> </em><em>image</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>caught</em><em> </em><em>on</em><em> </em><em>a</em><em> </em><em>screen</em></h3>
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em>
<em>will</em><em> </em><em>give</em><em> </em><em>the</em><em> </em><em>brainliest</em><em>!</em>
<em>follow</em><em> </em><em>~</em><em>H</em><em>i</em><em>1</em><em>3</em><em>1</em><em>5</em><em>~</em>
Answer:
B) the change in momentum.
Explanation:
The impulse is defined as the product between the force applied on an object (F) and the duration of the collision (
):
(1)
We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

So, (1) becomes

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time:
. If we substitute into the previous equation, we find

And the quantity
is equivalent to the change in momentum,
.
Answer:
Explanation: find the attached solution below
The bullet travels a horizontal distance of 276.5 m
The bullet is shot forward with a horizontal velocity
. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>
The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.
The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.
Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

Substitute 0 m/s for
, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

The horizontal distance traveled by the bullet is given by,

Substitute 500 m/s for
and 0.5530s for t.

The bullet travels a distance of 276.5 m.
Answer: 1026s, 17.1m
Explanation:
Given
COP of heat pump = 3.15
Mass of air, m = 1500kg
Initial temperature, T1 = 7°C
Final temperature, T2 = 22°C
Power of the heat pump, W = 5kW
The amount of heat needed to increase temperature in the house,
Q = mcΔT
Q = 1500 * 0.718 * (22 - 7)
Q = 1077 * 15
Q = 16155
Rate at which heat is supplied to the house is
Q' = COP * W
Q' = 3.15 * 5
Q' = 15.75
Time required to raise the temperature is
Δt = Q/Q'
Δt = 16155 / 15.75
Δt = 1025.7 s
Δt ~ 1026 s
Δt ~ 17.1 min