An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it
1 answer:
Answer:
v = 7.4 m/s
Explanation:
Given that,
Mass if a volleyball, m = 5 kg
The ball reaches a height of 2.8 m
We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :
So, the required speed is 7.4 m/s. Hence, the correct option is (b).
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Answer:
A. 69.9m
Explanation:
Given parameters:
Initial velocity = 10.5m/s
Final velocity = 21.7m/s
Time = 4.34s
Unknown:
Distance traveled = ?
Solution:
Let us first find the acceleration of the car;
Acceleration =
v is final velocity
u is initial velocity
t is the time
Acceleration = = 2.58m/s²
Distance traveled;
V² = U² + 2aS
21.7² = 10.5² + 2 x 2.58 x S
360.64 = 2 x 2.58 x S
S = 69.9m
Answer:
a) F₃₁ = 63.0 μN
b) F₃₂ = - 14.0 μN
c) q₂ = - 5.0 nC
Explanation:
a)
- Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
- So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:
b)
- Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:
c)
- Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows: