Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
Approximately
(assuming that the melting point of ice is
.)
Explanation:
Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

The energy required comes in three parts:
- Energy required to raise the temperature of that
of ice from
to
(the melting point of ice.) - Energy required to turn
of ice into water while temperature stayed constant. - Energy required to raise the temperature of that newly-formed
of water from
to
.
The following equation gives the amount of energy
required to raise the temperature of a sample of mass
and specific heat capacity
by
:
,
where
is the specific heat capacity of the material,
is the mass of the sample, and
is the change in the temperature of this sample.
For the first part of energy input,
whereas
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
Similarly, for the third part of energy input,
whereas
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
The second part of energy input requires a different equation. The energy
required to melt a sample of mass
and latent heat of fusion
is:
.
Apply this equation to find the size of the second part of energy input:
.
Find the sum of these three parts of energy:
.
It pushes the currents to opposite sides
Answer:
<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

- v is image distance
- u is object distance, u is 10 cm
- f is focal length, f is 5 cm

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>
• Let's derive this formula from the lens formula:

» Multiply throughout by fv

• But we know that, v/u is M

- v is image distance, v is 10 cm
- f is focal length, f is 5 cm
- M is magnification.

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>
- Image is magnified
- Image is erect or upright
- Image is inverted
- Image distance is identical to object distance.