Answer:
Jari
Explanation:
To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.
For Jari's path, starting point is (0, 0) and (6, 7) is another point.
The gradient is the difference in y divided by the difference in x:
Change in y=7-0=7
Change in x=6-0=6
Thus, the slope equals 7/6.
For Jade, her first point is (0, 10) and another is (6, 16).
Change in y=16-10=6
Change in x=6-0=6
Thus, the slope equals 6/6=1.
It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.
Answer:
Maximum emf = 5.32 V
Explanation:
Provided data includes:
Number of turns, N = 10
Radius of loop, r = 3 cm = 0.03 m
It made 60 revolutions each second
Magnetic field, B = 0.5 T
We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:
For the maximum emf, 
Therefore,
Hence, the maximum emf generated in the loop is 5.32 V.
<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack.
centripetal force = weight of the ball
m v^2 / r = m g
v^2 / r = g
v^2 = g r
v = sqrt { g r }
v = sqrt { (9.80~m/s^2) (0.7 m) }
v = 2.62 m/s
Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>
Answer:
F = 0.535 N
Explanation:
We will apply energy concepts, considering both the peak and the bottom of the path.
Top
Em₀ = U = mg y
Bottom
= K = ½ m v²
Emo =
mg y = ½ m v²
v = √ (2gy)
y = L - L cos θ
v = √ (2g L (1 - cos θ))
Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.
F = ma
a = v² / r
For the turning radius, the cable length is r = L.
F = m 2g (1 - cos θ)
Now, let's find the result.
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
