Answer:
Explanation:
Mass of the cable car, m = 5800 kg
It goes 260 m up a hill, along a slope of 
Therefore vertical elevation of the car = 
Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = 
). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.
Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.
Hence, total change in energy = mgh = 
where, g = acceleration due to gravity
h = height/vertical elevation
1 angstrom = 10^-8 cm
6.5 x 10-4 cm = 65 000 x 10-8 cm = 65 000 angstroms
answer 65 000 angstroms
Answer:
Option A
20 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision equals zero hence for this case also, the sum of momentum of the two cars before collision should be equal to momentum after collision.
Momentum is given as the product of mass and velocity, where velocity considers the direction. Momentum, p=mv where m is the mass of the object and v is the velocity.
Momentum before collision
Since the velocities of both cars before collision are zero hence their momentum are zero ie 0.5*0+(1*0)=0
Momentum after collision
For 0.5 Kg car, its momentum will be 0.5*-40=-20 Kg.m/s
For the 1 kg car its momentum will be 1*v=1v
v-20=0
v=20 m/s
Answer:
Explanation:
Let the initial rotational inertia be I and final rotational inertia be I / 6 .
Let the initial angular velocity be ω₁ and final angular velocity be ω₂.
Applying conservation of angular momentum law
I x ω₁ = I / 6 x ω₂
6 ω₁ = ω₂
initial rotational kinetic energy = 1/2 I x ω₁ ²
Final rotational kinetic energy = 1/2 ( I / 6 ) x ω₂ ²
Final rotational kinetic energy / initial rotational kinetic energy
= ( 1 / 6 ) x ω₂ ² / ω₁ ²
= ω₂ ² / 6ω₁ ²
= 36 ω₁ ² / 6ω₁ ²
= 6 .
