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soldier1979 [14.2K]
3 years ago
14

The teacher shared a demonstration with the same students. She begin by standing on a desk, holding a book and an open piece of

paper in the air, away from her body.. She asked her students: "Which will hit the ground first if I drop them?" Based on the previous data, the students answered that they would hit the ground at the same time. When she dropped them, the book hit the ground first. Next she crumpled up the paper and repeated the experiment.
Elaborate on the two paper drops. Was there different results? Why?

A) No. In both cases, the book dropped faster than the paper because it had so much more mass.
B) Yes. The crumpled paper dropped at the same rate as the book; the flat sheet did not due to air resistance.
C) In both cases the paper fell at the same rate because in both cases the air resistance and the masses were the same.
D) Yes there was a difference in the rate at which the papers fell. The crumpled paper had the same air resistance as the book so they fell at the same rate.

Physics
1 answer:
bazaltina [42]3 years ago
7 0

Answer:

I think its c

Explanation:

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xxMikexx [17]

Answer:

w=19.76 \ rad/s

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

w=w_o+\alpha \ t

Where \alpha is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

\displaystyle v_t=w.r

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call w_1 the angular speed of the loaded reel and w_2 that from the empty reel. We have

w_1=w_{o1}+\alpha_1 \ t

w_2=w_{o2}+\alpha_2 \ t

If r_f=35\ mm=0.035\ m is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

\displaystyle w_{01}=\frac{v_t}{r_f}

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

t=1.8\ h=1.8*3600=6480\ sec

\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s

\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s

If r_e=10\ mm=0.001\ m is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s

The full reel goes from w_{01} to w_{02} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}

\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2

The empty reel goes from w_{02} to w_{01} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2

So the equations for both reels are

w_1=1.098+0.00576 \ t

w_2=38.426-0.00576 \ t

They will be the same when

1.098+0.00576 \ t=38.426-0.00576 \ t

Solving for t

\displaystyle t=\frac{38.426-1.098}{0.0115}

t=3240 \ sec

The common angular speed is

w_1=1.098+0.00576 \ 3240=19.76 \ rad/s

w_2=38.426-0.00576 \ 3240=19.76 \ rad/s

They both result in the same, as expected

\boxed{w=19.76 \ rad/s}

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3 years ago
If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed
Alexus [3.1K]
120 minutes=2 hours
20/2= 10mph
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3 years ago
A rectangular piece of​ cardboard, whose area is 352 square​ centimeters, is made into an open box by cutting a 2​-centimeter sq
Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

xy=352

y=\dfrac{352}{x} -------- (1)

The volume of the rectangular piece

2(x-4)(\dfrac{352}{x}-4)=432

x^2-38x+352=0

(x-16)(x-22)=0

x=16,22

Put the value of x in the equation (I)

For x = 16

y=\dfrac{352}{16}=22

For x = 22

y=\dfrac{352}{22}=16

Dimension of cardboard is 22 m by 16 m

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A, hope this helped! I didn’t really get it but I think it’s correct?
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3 years ago
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