This specific question applies the concept of relative velocity. Relative velocity is describes as the velocity of an object with respect to another object, whether static or moving. For this problem, the correct statement is t<span>he spotter above the tunnel will observe the speed of the spy as 95 km/h, and the person in the car will observe the speed of the spy as 200 km/h. The following are the justification, relative speed of the spotter at the tunnel is equal to velocity (Vss) = velocity spy relative to the ground (Vsg) + V ground relative to the spotter (Vgs). Thus, V = 95 + 0 km/h = 95 km/h. Note that the sign of velocity is dependent on the direction, hence, opposite direction is equal to negative velocity. For the second condition, velocity of the spy in reference person in the car is equal to velocity of the spy reference to ground plus the velocity of the ground reference to the person in the car. Thus, V = 95 km/h + 105 km/h = 200 km/h. Please be cautious in assigning the sign of the velocity.</span>
The elastic potential energy of a spring can be said to be the energy possessed by a spring due to a force that compresses or stretches the spring. This can also be said to be the work which is done by the applied force against the restoring force of the spring, which is stored as the elastic potential energy.
The elastic potential of a spring can be expressed as:
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Answer:
1 l = 1000 cc so 1 ml = 1 cc
this is the volume
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Answer:
185.25 m/s
Explanation:
consider the motion of the combination of bullet and block after the collision
v₀ = initial speed just after the collision
v' = final speed = 0 m/s
μ = Coefficient of friction = 0.6
g = acceleration due to gravity = 9.8 m/s²
a = acceleration of the combination = - μ g = - (0.6) (9.8) = - 5.88 m/s²
d = stopping distance = 13 m
using the kinematics equation
v'² = v₀² + 2 a d
0² = v₀² + 2 (- 5.88) (13)
v₀ = 12.4 m/s
m = mass of the bullet = 9.9 g = 0.0099 kg
M = mass of the wood = 138 g = 0.138 kg
v = speed of bullet before collision
v₀ = speed of combination after the collision = 12.4 m/s
Using conservation of momentum
m v = (m + M) v₀
(0.0099) v = (0.0099 + 0.138) (12.4)
v = 185.25 m/s