Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
Amount of work done is zero and so power = 0 watts.
<u>Explanation:</u>
Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.
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Answer:
θ = 8.50°
To the nearest angle
θ = 9.0°
the golfer must hit the ball at angle 9° so that it travels 120 feet.
Explanation:
The range of a projectile is the horizontal distance covered by a projectile, which can be written as;
r = (u^2× sin2θ)/g
Where;
r = range
u = initial speed
θ = angle from horizontal
g = acceleration due to gravity
Solving for θ,
sin2θ = rg/u^2
θ = 1/2 × sin⁻¹(rg/u^2) ....1
Given;
r = 120 ft
u = 115 ft/s
g = 9.81m/s = 32.2 ft/s
Substituting the values into the equation 1;
θ = 1/2 × sin⁻¹(120×32.2/115^2)
θ = 1/2 × sin⁻¹(0.29217)
θ = 1/2 × 17.00
θ = 8.50°
To the nearest angle
θ = 9.0°
