Answer:
For C is B and for D is A
Explanation:
B because that's where the b car will do all the force to go up, and A because there the car doesn't have to do any force other than the natural, and it already brings that on it's own.
Answer:
The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.
Explanation:
Given that,
Mass of softball = 0.220 kg
Speed = 5.5 m/s
(a). We need to calculate the velocity of the target ball
Using conservation of momentum
....(I)
The velocity approach is equal to the separation of velocity
(b). We need to calculate the mass of the target ball
Now, Put the value of v₂ in equation (I)
Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.
Answer:
1.) 1620 km/h^2
2.) 2.7 km
Explanation:
1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,
U = 0.
Final velocity V = 54 km/h
Time t = 2 minute = 2/60 = 1/30 hour
Acceleration a will be change in velocity per time taken. That is,
a = ( V - U )/ t
Substitute V, U and t into the formula
a = 54 ÷ 1/30
a = 54 × 30 = 1620 km/h ^2
2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.
S = Ut + 1/2at^2
Substitute all the parameters into the formula
S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2
S = 54/30 + 810 × 1/900
S = 54/30 + 810/900
S = (1620+810)/900
S = 2430/900
S = 2.7 km.
Therefore, distance travelled by the car during this time is 2.7 km
Answer:C
Explanation:the both are telling us more about a gasket
