When fluorine reacts, it will gain electron(s). O 3 O 1 ОО O2

Aleksandr [31]

Let's cut through the weeds and the trash
and get down to the real situation:

A stone is tossed straight up at 5.89 m/s .
      Ignore air resistance.

Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

   (5.89 m/s / 9.8 m/s²) = 0.6 seconds.

At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at

(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.

   (0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.

5 0

4 years ago

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration of a = 2.20 m/s2. A worker assists

trasher [3.6K]

Answer:

Explanation:

Given

acceleration a =2.2 m/s^2

Force F(t) is given by

F(t)=5.40 t N/s

Power supplied by this force at different time t is given by

Power $=F\cdot v$

velocity at any instant t is given by

v=at

Power $=5.40 t\cdot 2.2 t$

Power $=11.88 t^2$

at t=1 s

Power =11.88 W

at t=2 s

Power $=11.88\times 4=47.52$

at =4 s

Power $=11.88 \times 16=190.08$

6 0

3 years ago

Please answer this as soon as possible

Alinara [238K]

I think it’s the third one, if not it’s the first.

4 0

3 years ago

When your vehicle is properly parked in a straight-in parking space,

marin [14]

The correct answer to this question would be:

A. No part of your vehicle will extend out into the traffic lane.

This kind of maneuver only shows your skill to handle the vehicle in tight spaces, ability to judge distance, and showing control of the vehicle as you turn into a straight-in parking space.

3 0

3 years ago

Read 2 more answers

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

#SPJ4

3 0

1 year ago