#1). Anthony does the same amount of work as Angel, with <em>more power</em>.
#2). Power = (Work)/(Time) = 41,000 J / 500 s = <em>82 watts .</em>
#3). Power = (Work) / (Time) = 83 J / 3 sec = <em>27.7 watts</em>
The reference point that should be used to describe Melanie motion is the pond (option C).
<h3>What is reference point?</h3>
Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.
According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.
However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.
Learn more about reference point at: brainly.com/question/14318992
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We are given with the x and y components of Vector A and B. In this case, we compute the resultant of both components of each vector. The vector is equal to the square root of the sum of the squares of the components. A is equal to 2.60 cm. B is equal to 5.56 cm. B is found in quadrant Iv and has an angle of 42.447 degrees as a terminal angle. A has an angle of 59.98 degrees.
a. 5.6082 < -15.53 degreesc. 6.63 <-64.98 degreesb. x = 6.63 cos -64.98 degrees = 2.80 y = 6.63 sin -64.98 degrees = -6.00
Answer:
4 m/s²
Explanation:
When the elevator is 1 m below point of contact , compression will be 1 m.
Restoring force in the spring will be 10600 N. Friction force of 17000N will also act in upward direction . The weight of 2000 x 9.8 N will act downwards
Force in down ward direction = 2000 x 9.8
= 19600 N
Force in upward direction
= 10600 + 17000
= 27600 N
Net force in upward direction
= 27600 - 19600
= 8000 N
Acceleration in upward direction
= 8000 / 2000
= 4 m/s²
Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C = 
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C = 
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ = 
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
