Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Answer:
3.59 m/s
Explanation:
We are given that
Mass of lineman,m=85 kg
Mass of receiver,m'=90 kg
Speed of receiver,v'=5.8 m/s
Speed of lineman,v=4.1 m/s
We have to find the their velocity immediately after the tackle.
Initial momentum,
According to law of conservation of momentum
Initial momentum=Final momentum=
Answer:
Explanation:
The process during which pressure remains constant is called an isobaric process.
here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.
So angle must be more than 270 degree and less than 360 degree
Now in order to find the value we can say that
so it is inclined at above angle with X axis in fourth quadrant
Now if angle is to be measured counterclockwise then its magnitude will be
so the correct answer will be 305 degree