Answer:

Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,

v = 2v

..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,


So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Answer:
correct is d) a ’= g / 2
Explanation:
For this exercise let's use the kinematics equations
On earth
v = v₀ - a t
a = (v₀- v) / T
On planet X
v = v₀ - a' t’
a ’= (v₀-v) / 2T
Let's substitute the land values in plot X
a’= a / 2
Now let's use Newton's second law
W = ma
m g = m a
a = g
We substitute
a ’= g / 2
So we see that on planet X the acceleration is half the acceleration of Earth's gravity
Answer:
The specific heat is 3.47222 J/kg°C.
Explanation:
Given that,
Temperature = 13°C
Temperature = 37°C
Mass = 60 Kg
Energy = 5000 J
We need to calculate the specific heat
Using formula of energy


Put the value into the formula


Hence, The specific heat is 3.47222 J/kg°C.