The correct answer is 19.72 °C. The first step is to determine the amount of heat that was lost per gram (30,000 J) / (390 g) = 76.92 J/g. Then to determine the temperature change, divide 76.92 J/g with the given specific heat of milk (76.92 J/g) / (3.9 J/g°C) = 19.72 °C.
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
answer:
yes
explanation:
At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge 
at a point 
distance away is given by
where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge .
= unit vector along .
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, 
are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
