Answer:
75 km/s^2
Explanation:
Divide the final velocity by the time needed to reach the final velocity.
(1500 m/s) /.02s = 75,000 m/s^2 = 75 km/s^2
Answer:
Consider the followig calculation
Explanation:
a) use deal equation:
PV = nRT
ρ = m/V,= ==> V = m/ρ
therefore,
ρ = Pm/RT
convert 95 oF in degree
95 oF = 308.15 K
1 atm = 1.013 * 105 pascal
ρ = 1.013*105 * 29 * /8.314 * 308.15
= 1.146 kg/m3
b) again use ideal gas equation:
ρ = Pm/RT
T = 50 oF = 283.15 K
1 atm = 1.013 * 105 pascal
molar mass will be same
ρ = 1.013 * 105 * 29 / 8.314 * 283.15
ρ = 1.248 kg / m3
So,
c) . more than density of the hot, dry air computed in part (a)
Answer:
We know that the torque can be calculated as follows:
T = rpsinα
With r being the distance of the body from the center of the circumference he has as trajectory, p being the momentum of the body and sinα being the sine of the angle between the 2 vectors: r and p.
It's pretty obvious that T is directly proportional to the momentum, that can be written as p = m·v, with m being the mass of the object and v the velocity of the object.
Answer:
A. the magnitude of the velocity at which the two players move together immediately after the collision is 7.9m/s
B. The direction of this velocity is due north as the linebacker since he has obviously has more momentum
Explanation:
This problem bothers on the inelastic collision
Given data
Mass of linebacker m1= 110kg
Mass of halfbacker m2= 85kg
Velocity of linebacker v1= 8.8m/s
Velocity of halfbacker v2= 7.2m/s
Applying the principle of conservation of momentum for inelastic collision we have
m1v1 +m2v2= (m1+m2)v
Where v is the common velocity after impact
Substituting our data into the expression we have
110*8.5+85*7.2= (110+85)v
935+612=195v
1547=195v
v=1547/195
v=7.9m/s
Momentum of linebacker after impact = 110*7.9= 869Ns
Momentum of halfbacker after impact = 85*7.9= 671.5Ns
the direction after impact is due north since the linebacker has greater momentum
