Answer:
The correct option is;
B) The specific heat of ice is less than that of water.
Explanation:
Here we have
Let the amount of energy added to the ice at -10 C to raise the temperature to -5 C be X J
Let the amount of energy added to the water at 15 C to raise the temperature to 20 C be Y J
We know that the heat required, ΔQ to raise the temperature of a substance is given by
ΔQ = m·c·Δθ
Where:
m = Mass of the substance
c = Specific heat capacity
Δθ = Temperature change
Since the mass of the ice and the water are the same, so also is the change in temperature, (-5 - (-10) = 5 and 20 - 15 = 5) we have
for m₂·c₂·Δθ₂ > m₁·c₁·Δθ₁
Where:
m₁, c₁, Δθ₁, is for the ice and m₂, c₂, Δθ₂ is for the water and
m₁ = m₂
Δθ₁ = Δθ₂
Therefore,
c₂ > c₁ = c₁ < c₂
That is the specific heat capacity of the ice is lesser than the specific heat capacity of the water.
The kinetic energy increase because as temperature warms up particles move faster
Answer:
A = (27.95 N, 21 N)
Explanation:
The polar co-ordinates are given as:
(r,θ) = (35 N, 37°)
Now, to convert this into polar co-ordinates (x, y), we will use following relations:
r² = x² + y²
(35)² = x² + y²
1225 = x² + y² ----------- equation (1)
and
tan θ = y/x
tan 37° = y/x
y = 0.753 x ------------------- equation (2)
Substituting this value in equation (1):
1225 = x² + (0.753 x)²
1225 = 1.567 x²
x² = 1225/1.567
x = √781.32
x = 27.95 N
using this value in equation (2)
y = (0.753)(27.95 N)
y = 21 N
Therefore, the vector can be represented in polar co-ordinates as:
<u>A = (27.95 N, 21 N)</u>
Answer:
hey I know the answer for this
Explanation:
it will be reduce the speed of A
the answer is 7.
this is the exact answer.
if you don't like this answer means pls comment me
