Answer:
Depends on where he was fouled
If it is beyond the 3 pt line, then it is 3
If it is within the 2 pt line, then it is 2
I hope that helps :)
Explanation:
Plz mark B R A I N L I E S T
The total energy of a particle is its rest energy plus the kinetic energy. Its formula is: Et= m^2/sqrt((1-v^2)/c^2))
The rest energy is equal to the product of mass and the square of light velocity: Er=mc^2.
When the kinetics energy is twice its rest energy this holds:
Et=Er
m^2/sqrt((1-v^2)/c^2))=<span>mc^2.
</span><span>sqrt((1-v^2)/c^2))=m/c^2
</span>=> v=sqrt(3/2)c
Answer:
F = mv²/(18d)
7
Explanation:
Momentum before collision = momentum after collision
mv = m(v/3) + (4m) V
v = v/3 + 4V
4V = 2v/3
V = v/6
The acceleration of the block is:
v² = v₀² + 2aΔx
(0 m/s)² = (v/6)² + 2ad
2ad = -v²/36
a = -v²/(72d)
The friction force is therefore:
∑F = ma
-F = (4m) (-v²/(72d))
F = mv²/(18d)
The energy dissipated during the collision is:
ΔE = 1/2 mv² − (1/2 m(v/3)² + 1/2 (4m)(v/6)²)
ΔE = 1/2 mv² − (1/18 mv² + 1/18 mv²)
ΔE = 1/2 mv² − 1/9 mv²
ΔE = 7/18 mv²
The energy dissipated by the friction force:
W = Fd
W = 1/18 mv²
The ratio is:
ΔE / W
(7/18 mv²) / (1/18 mv²)
7
The force of friction is 25 N
Explanation:
For this problem, we can apply Newton's second law of motion along the horizontal direction:
where
is the net force in the horizontal direction
m is the mass of the block
is the horizontal acceleration
Here the block is moving at constant speed, so its acceleration is zero, therefore:
(1)
The net force in the horizontal direction can be written as:
(2)
where
is the horizontal component of the pulling force, with F = 50 N being the magnitude and 
being the direction, acting forward
is the force of friction, acting backward
Combining (1) and (2), we find the magnitude of the force of friction:
Learn more about friction:
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