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Shalnov [3]
3 years ago
6

A 55 kilogram person jumps off a cliff and hits the water 5.8 seconds later, how high is the cliff above the water?

Physics
1 answer:
dusya [7]3 years ago
5 0
The cliff is 319. kilograms above the water
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Einstein’s general theory of relativity made or allowed us to make predictions about the outcome of several experiments that had
PtichkaEL [24]

Answer:

1. bending of light in gravitational fields.

2. effect of gravitational redshift.

3. perihelion precission of mecury.

Explanation:

1 bending of light in gravitational fields, we can think of it like this:

by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.

2. effects of gravitational redshift:

this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.

3. perihelion precission of mecury:

according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.

3 0
4 years ago
I have my exam monday I need some help!
Lana71 [14]

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

f=\frac{v}{\lambda} and filling in:

2.5=\frac{v}{2.0} so

v = 2.5(2.0) and

v = 5.0 m/s

7 0
3 years ago
Why forces are balanced and unbalanced? need help with this the lesson is tommorow
Naya [18.7K]
"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way. 

-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down.  It's just as if there's nobody sitting on it.

-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move.  The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.

-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her.  If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.

From these examples, you can see a few things:

-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.

-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.

-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
4 0
3 years ago
Read 2 more answers
What is the average speed of a kangaroo that hops 60 m in 5 s ?
jok3333 [9.3K]
The formula for average speed is
v_{avg}= \frac{total distance}{total time}
So we can just substitute our data.
v_{avg}= \frac{60}{5}=12 \frac{m}{s} - its the result
6 0
3 years ago
Read 2 more answers
A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
Bingel [31]

Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
4 years ago
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