The answer is letter a. It is best to slow down in situations of heavy rain or flooded road as skid could be the result if you lose out of control because the driver isn't slowing down. That is why it is being said that tires can ride on a thin film of water skis as it could skid if it has lost control if the driver hadn't slowed down.
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s
The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²
The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j
The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J
The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J
Answer: 0.0226 J
The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:
where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
Learn more about vector addition:
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Answer:
15,505 N
Explanation:
Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student
-ΔU = ΔK
-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²
-(0 - mgh) = 1/2mv² - 0
mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.
mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 1 m)
v = √(19.6 m²/s²)
v = 4.43 m/s
Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s
So Ft = m(v' - v)
F = m(v' - v)/t
substituting the values of the variables, we have
F = 70 kg(0 m/s - 4.43 m/s)/0.02 s
= 70 kg(- 4.43 m/s)/0.02 s
= -310.1 kgm/s ÷ 0.02 s
= -15,505 N
So, the force transmitted to her bones is 15,505 N
