Answer:
B. 1500 kg*m/s
Explanation:
Momentum p = m* v
In any type of collision, the total momentum is preserved!
The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.
p1 + p2 =
m1*v1 + m2*v2
m1 = me = 300 kg
v1 = 3 m/s
v2 = 2 m/s
Substitute the given numbers:
300*3 + 300+2
900 + 600
1500 kg*m/s, which is answer B.
Answer:
Explanation:
A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.
In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.
When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.
Answer:
The potential energy increases and the kinetic energy decreases
Answer:
42.6 m
Explanation:
mass of crate m = 53 kg
coefficient of kinetic friction, μ = 0.36
acceleration due to gravity, g = 9.8 m/s^2
Force, F = 372.098 N
Net force, f = F - friction force
f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8
f = 185.114 N
acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2
initial velocity, u = 0
time, t = 4.94 s
s = ut + 1/2 at^2
s = 0 + 1/2 x 3.49 x 4.94 x 4.94
s = 42.6 m
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
