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Lady_Fox [76]
3 years ago
13

A 140

Physics
1 answer:
Doss [256]3 years ago
7 0
There is not enough information given tl be able to solve this problem
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CAN SOMEONE DO THIS TO ME ITS OK IF YOUR DRAWING IS NOT THAT COOL OR GOOD AND DONT GET A PHOTO FROM SOCIAL MEDIA
Svet_ta [14]

bc i cant draw good sorry but hope it helps!

and can i have brainliest? ight plz can i have brainliest?

ill give another picture of the volley ball thing

is that better?

7 0
3 years ago
When analyzing a position time graph the direction of the slope positive or negative tells us:
Sphinxa [80]

Answer:

Explanation:

Positive values for position indicate that the object is in front of the starting point and negative values tell us that the object is behind the starting point. (time = 9.5, position = 0) the object is at the starting point.

8 0
3 years ago
Read 2 more answers
What is the force required to lift the balloon with
Olin [163]

No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.

Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !

3 0
3 years ago
A 0.350kg bead slides on a curved fritionless wire,
LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

m1*g*h-\frac{m1*V_B^2}{2}=0  Solving for Vb:

V_B=\sqrt{2gh}=6.56658m/s

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

m1*g*h2-\frac{m1*V_{B'}^2}{2}=0 Solving for h2:

h2 = 0.092m

6 0
3 years ago
How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?
tangare [24]

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon !   If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove.  It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels.  Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up.  We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."

3 0
3 years ago
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