it moves toward the truck because increased air movement between the car and the truck decreases pressure.
Hope this helped :) <3
Answer:
A. 1.4 m/s to the left
Explanation:
To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:
where:
M = momentum [kg*m/s]
M = m*v
where:
m = mass [kg]
v = velocity [m/s]
where:
m1 = mass of the basketball = 0.5 [kg]
v1 = velocity of the basketball before the collision = 5 [m/s]
m2 = mass of the tennis ball = 0.05 [kg]
v2 = velocity of the tennis ball before the collision = - 30 [m/s]
v3 = velocity of the basketball after the collision [m/s]
v4 = velocity of the tennis ball after the collision = 34 [m/s]
Now replacing and solving:
(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)
1 - (0.05*34) = 0.5*v3
- 0.7 = 0.5*v
v = - 1.4 [m/s]
The negative sign means that the movement is towards left
Answer:
You drop a rock from rest out of a window on the top floor of a building, 30.0 m above the ground. When the rock has fallen 3.00 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the ...... rest out of a window on the top floor of a building, 30.0m above the ground. ... You Notice That Both Rocks Reach The Ground At The Exact Same Time. ... You drop a rock from rest out of a window on the top floor of a building, 30.0m ... When the rock has fallen 3.20 m, your friend throws a second rock straight down from ...
Answer:An object can be considered a point object if during motion in a given time it covers a distance much greater than its own size.
Explanation:
The ocean does not change temperature but it does lose some entropy ( he gives heat to melt the ice and to warm it to 3.70° C ).
I ) For the ice:
1 ) For melting the ice:
Q = m · Lf = 4.45 kg · 334 · 10³ J/kg = 1,486,300 J
Δ S = Q / T = 1,486,300 J / 273.15 K = 5.441 · 10³ J/K.
2 ) To warm the melted ice to 3.70° C:
Q = m c Δ T = 4.45 kg · 4,190 J / kgK · 3.70 K = 68,988.35 J
Δ S = m c ln( T2/ T1 ) = 4.45 kg · 4,190 J/kgK · ln ( 276.85 / 273.15 ) =
= 18,645.5 · ln ( 1.01354 ) = 18.645.5 · 0.013454 = 250.8692 J/K
II ) For the ocean:
Δ S = Q / T = ( - 68,988.35 + 1,486,300 ) / 276.85 = - 5,617.8 J/K
The net entropy change:
Δ S = ΔS ice + ΔS ocean = 5,441.1 + 250.8692 - 5,617.8 = 74.1692 J/K
Answer: 74 J/K.
