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Nutka1998 [239]
3 years ago
12

At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta

nt) speed vB = 52 m/s. The airplane has the same constant speed vA = 52 m/s, and after a period of level flight is just beginning to follow the circular path shown of radius ρA = 2330 m. (a) Determine the velocity and acceleration of the airplane relative to the skydiver. (b) Determine the time rate of change of the speed vr of the airplane and the radius of curvature ρr of its path, both as observed by the nonrotating skydiver.
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer:

a=2330

b= 0.223secs

Explanation:

pb=2330m

t=0.223secs

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Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?
Helen [10]

Answer:

A. Mass and acceleration

Explanation:

  • According to Newton's second law of motion, the resultant force is directly proportional to the rate of change in momentum
  • Therefore; F = ma , where F is the resultant force, m is the mass, and a is the acceleration of the body.
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3 0
3 years ago
Read 2 more answers
A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol
olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa,  т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P )  = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅  ,   where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅  of  CD on the shaft BC

Torque = 660 * 0.3 cos∅

             = 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0
3 years ago
The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the
vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r²             where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

4 0
3 years ago
A spider twirls a 25 mg fruit fly around in a circle with radius 17.6 cm at the end of a web. If the velocity of the fly is 110
olchik [2.2K]

Answer:

Fc =  1.7x10^-4 N

Explanation:

Convert everything to proper units:

m = 25mg = 2.5x10^-5 kg

r = 17.6cm = 0.176m

v = 110cm/s = 1.1m/s

the formula for centripetal force is Fc = mv^2 / r

Plug everything and solve for Fc;

fc = (2.5x10^-5)(1.1^2) / 0.176

Fc =  1.7x10^-4 N

8 0
4 years ago
A balloon with a charge of 6.0 μC is held a distance of 0.80 m from a second balloon having the same charge. Calculate the magni
Yuliya22 [10]

Answer:0.506 N

Explanation:

Given

Charge on first balloon q_1=6\ \mu C

Charge on second balloon is q_2=6\ \mu C

Distance between them r=0.8\ m

Electrostatic Repulsive force is given by

F=\dfrac{kq_1q_2}{r^2}

Where K is constant

F=\dfrac{9\times 10^9\times 6\times 10^{-6}\times 6\times 10^{-6}}{(0.8)^2}

F=\dfrac{324\times 10^{-3}}{0.64}

F=0.506\ N

6 0
3 years ago
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