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VARVARA [1.3K]
3 years ago
12

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

e-half his normal weight. How fast is he going
Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

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C. It helps scientists detect dark matter.

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3 years ago
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A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of
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3 years ago
A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d
zvonat [6]

Answer:

<u />D_l=d<u />

Explanation:

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The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

V_{e1}=V_{e2}

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Where

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And

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5 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

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3 years ago
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4 x 10^-9F

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