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Maru [420]
4 years ago
8

A squeeze bottle squeezes when pressed. It regains its shape when pressed. It regains its shape when pressure from your hand wit

hdrawn. What may happen if squeeze bottle is pressed to take sauce out and then immediately corked tightly? Will it regain its shape? If not WHY?
Physics
1 answer:
Vaselesa [24]4 years ago
6 0

Answer:

the bottle would stay in its squashed shape, and it will only regain its shape if u remove the cork

Explanation:

because theres not enough air inside to let the bottle to go back its normal form. Its like when u let the air out of a balloon, theres no air inside to let it stay in its 'big or expanded' form.

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A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay
yuradex [85]

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = 20\times 1=20\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2

v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

4 0
4 years ago
Which of these is true about kinetic energy but not necessarily true about potential energy
cestrela7 [59]

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference.  So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative. 

What that means is:  You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom.  Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes.  So is potential energy.

-- Kinetic energy increases with height ...
   No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
   No. It doesn't, but potential energy does.

3 0
3 years ago
a diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Diver has a speed of 1.2m/s.
8090 [49]
When the diver reaches maximum height, the upward velocity will be zero.

We shall use the formula
v^2 = u^2 - 2gh
where 
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.

Therefore
0 = 1.2^2 - 2*9.8*h
h = 1.2^2/(2*9.8) = 0.0735 m

Answer: 0.074 m (nearest thousandth)
5 0
3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
3 years ago
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Kay [80]
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5 0
3 years ago
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