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allsm [11]
2 years ago
13

You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque

s on you. Earth’s gravity is effectively pulling downward on you at your center of gravity. Consider a net force and a net torque about your center of mass. Which of these two influences are acting on you as you fall?
a. The net force is zero, but the net torque is nonzero.
b. The net force is nonzero, but the net torque is zero.
c. The net force is zero and the net torque is zero as well.
d. Both the net force and the net torque are nonzero.
Physics
1 answer:
harina [27]2 years ago
4 0

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

                 Fe - W = m a  

                 k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

                Fe y - W y = I α

angular and linear acceleration are related

       a = α r

       Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

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Answer:

Explanation:

Part A) Using

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A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B)  Using

I=c*ε*E^2/2

rearrange to solve for E= \sqrt{((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

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amplitude of the electric field E= \sqrt{(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= \sqrt{(1.8*10^11) / (2.66*10^-3) = \sqrt{(6.8*10^13) = 8.25*10^6 V/m    

 

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What is the energy range (in joules) of photons of wavelength 410 nm to 750 nm ? Express your answers using two significant figu
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Answer:

4.9 x 10^-19 J, 2.7 x 10^-19 J

Explanation:

first wavelength, λ1 = 410 nm = 410 x 10^-9 m

Second wavelength, λ2 = 750 nm = 750 x 10^-9 m

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E2 = 2.7 x 10^-19 J

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