25.82 m/s
Explanation:
Given:
Force applied by the baseball player; F = 100 N
Distance the ball travels; d = 0.5 m
Mass of the ball; m = 0.15 kg
To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved.
It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows:
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
Solving gives:
v² = (2 × 100 × 0.5) / 0.15
v² = 666.67
v = √666.67
v = 25.82 m/s.
If my calculations are accurate, the angle is 67.5 degrees.
Response:
(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v
Clarification:
Solution
Given that:
A refrigerator magnet with a depth of approximately 2 mm
The estimated magnetic field strength of the magnet is = 5 m T
The Area = 8 cm²
Now,
(A) The magnetic flux ΦB = BA
Therefore,
ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²
Thus,
ΦB = 4* 6 ^ ⁻6 T m²
(B) By employing Faraday's Law, the subsequent equation applies:
Ε = Bℓυ
Where,
ℓ = 2 cm equals 2 * 10 ^⁻2 m
B = 5 m T = 5 * 10 ^ ⁻3 T
υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s
Therefore,
Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v
E =2 * 10 ^ ⁻6 v