1 m/s
Explanation:
To solve this question we use the following formula:
momentum = mass × velocity
momentum of the first car = 1000 kg × 2.5 m/s
momentum of the second car = 2500 kg × X m/s
To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.
momentum of the first car = momentum of the second car
1000 kg × 25 m/s = 2500 kg × X m/s
X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s
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momentum
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Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
Hope this helps!
Answer:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N
Answer:
The forms of energy involved are
1. Kinetic energy
2. Potential energy
Explanation:
The system consists of a ball initially at rest. The ball is pulled down from its equilibrium position (this builds up its potential energy) and then released. The released ball oscillates due to a continuous transition between kinetic and potential energy.
