Answer:
The answer is 6.40 meters.
Explanation:
The speed v = √(2gh)
v = √( 2×9.8×6.4) = 11.2 m/s
After, finding the time it takes to hit the ground from a height of 1.6 meters.
time = √(2H÷g)
time = √(2×1.6÷9.8)
time = 0.5714 seconds.
Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.
W =F triangle d cosine0. F = 25 Newton’s. Delta d = 50 meters. Theta =40.0 degrees
Answer:
The maximum emf induced in the ring
= (2.882 × 10⁻⁷) V
Explanation:
According to the law of electromagnetic induction, the emf induced in the ring is given by
E = N BA w sin wt
The maximum emf induced is
E = N BA w
B = 30.5 μT = (30.5 × 10⁻⁶) T
A = (πD²/4)
D = 1.75 cm = 0.0175 m
A = (π×0.0175²/4) = 0.000240625 m²
Nw = 2π × 6.25 = 39.29 rad/s
E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29
E = (2.882 × 10⁻⁷) V
Hope this Helps!!!
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
The weight should be shared between the two string equally. Therefore, tension in each string, T is;
T = 120 N/2 = 60 N