Answer:
1. increases
2. increases
3. increases
Explanation:
Part 1:
First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:
F1 - fs = 0.
And this friction force fs is:
fs = Nμs,
where μs is the static coefficient of friction, and N is the normal force.
Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:
N = mg + F2.
So, F2 is increasing, that means fs is increasing too.
Part 2:
As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.
Part 3:
In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.
Its really hurts
Explanation:
Charge A and charge B are 2.2 m apart. Charge A is 1.0 C, and charge B is
2.0 C. Charge C, which is 2.0 C, is located between them and is in
electrostatic equilibrium. How far from charge A is charge C?
Answer:
Explanation:
mass of car, m = 1022 kg
mass of truck, M = 1620 kg
initial velocity of truck, U = 14.5 m/s
initial velocity of car, u = 0 m/s
Let the final velocity of car is v and the final velocity of truck is V.
Collision is elastic, so the coefficient of restitution, e = 1
Use conservation of momentum
initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck
m x u + M x U = m x v + M x V
0 + 1620 x 14.5 = 1022 v + 1620 V
23490 = 1022 v + 1620 V ..... (1)
Use the formula of coefficient of restitution
1 (14.5 - 0) = v - V
14.5 = v - V
V = v - 14.5 .... (2)
Put in equation (1)
23490 = 1022 v + 1620 (v - 14.5)
23490 = 1022 v + 1620 v - 23490
46980 = 2642 v
v = 17.8 m/s
Put in equation (2)
V = 17.8 - 14.5
V = 3.3 m/s
Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.
Answer:
The centre of the earth is harder to study than the centre of the sun." Temperatures in the lower mantle the reach around 3,000-3,500 degrees Celsius and the barometer reads about 125 gigapascals, about one and a quarter million times atmospheric pressure.
Explanation:
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
