Answer:
Force on front axle = 6392.85 N
Force on rear axle = 8616.45 N
Explanation:
As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels
Now we know that



now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle
so we can write torque balance about its center of mass


now from above equation

now we have

now the other force is given as

Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.
Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)
The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)
Answer: 1.1x10¹⁶ kg
Answer: 0.56 m/s
Explanation:
hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.
so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.
Now, the X component is definitely conserved here.
so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:

Answer:

Explanation:
For the first ball, the moment of inertia and the kinetic energy is:


So, replacing, we get that:

At the same way, the moment of inertia and kinetic energy for second ball is:


So:

Then,
is equal to
, so:




Finally, solving for
, we get:

Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s