Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
<span>Make the surfaces smoother. Rough surfaces produce more friction and smooth surfaces reduce friction
Lubrication is another way to make a surface smoother
Make the object more streamlined
Reduce the forces acting on the surfaces
<span>Reduce the contact between the surfaces.</span></span>
Angular velocity of the rotating tires can be calculated using the formula,
v=ω×r
Here, v is the velocity of the tires = 32 m/s
r is the radius of the tires= 0.42 m
ω is the angular velocity
Substituting the values we get,
32=ω×0.42
ω= 32/0.42 = 76.19 rad/s
= 76.19×
revolution per min
=728 rpm
Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
<u>τ = 4900 N.m</u>
<u></u>
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
<u>τ = 4900 N.m</u>
<u></u>
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
<u>τ = 4900 N.m</u>
<u></u>
Hence, the correct answer will be:
<u>e. The torque is the same for all cases.</u>
Answer:
a) K = 2/3 π G m ρ R₁³ / R₂
, b) U = - G m M / r
Explanation:
The law of universal gravitation is
F = G m M / r²
Part A
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / R₂
G m M / R₂² = m v² / R₂
v² = G M / R₂
They give us the density of the planet
ρ = M / V
V = 4/3 π R₁³
M = ρ V
M = ρ 4/3 π R₁³
v² = 4/3 π G ρ R₁³ / R₂
K = ½ m v²
K = ½ m (4/3 π G ρ R₁³ / R₂)
K = 2/3 π G m ρ R₁³ / R₂
Part B
Potential energy and strength are related
F = - dU / dr
∫ dU = - ∫ F. dr
The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1
U- U₀ = G m M ∫ dr / r²
U - U₀ = G m M (- r⁻¹)
We evaluate for
U - U₀ = -G m M (1 /
- 1 /
)
They indicate that for ri = ∞ U₀ = 0
U = - G m M / r