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Nuetrik [128]
4 years ago
11

According to Newton's third law, if you push on a wall, the wall ?

Physics
2 answers:
LenKa [72]4 years ago
5 0
The wall will push back.
Lady bird [3.3K]4 years ago
3 0

The wall will push back, in exactly the opposite direction, and with
exactly the same size force.

That's why the net force on the palm of your hand is zero, and that
in turn is the reason that your hand doesn't accelerate.

If you keep increasing the strength of your push, then eventually you
exceed the force that the wall is capable of delivering.  Then the wall
crumbles and falls, your hand accelerates in the direction you're pushing,
and the crowd goes wild !


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A wrecking ball has a mass of 315 kg. If it is moving at a speed of 5.12 m/s, what is its kinetic energy?
zubka84 [21]
The formula for Kinetic energy is:
KE (Kinetic Energy) = 1/2 M (mass) x V (velocity) ^2

Substitute the known values into the equation:
KE = 1/2 (315) x (5.12)^2

Find out the values in sections (split the formula in half)
1/2 x 315 = 157.5
5.12^2 = 26.2144

Now times the answers together to complete the formula.
157.5 x 26.2144 = 4,128.768 km^2/s
5 0
3 years ago
If a 9 V battery with 4 Ω contact resistance is used and the relay has 80 Ω and the wire has 10 Ω/mile, what is the maximum tele
Sedaia [141]

Answer:

Telegraph distance will be 9.6 mile

Explanation:

We have given the voltage V = 9 volt

Contact resistance = 4 ohm

Relay resistance = 80 ohm

And wire has a resistance = 10 ohm/mile

We have given the current = 50 mA =0.05 A

According to ohm's law R=\frac{V}{I}=\frac{9}{0.05}=180ohm

So the resistance of wire = 180-4-80=96 ohm

So the length of the telegraph distance will be \frac{96}{10}=9.6mile

4 0
4 years ago
Write any five uses of uv rays​
Elan Coil [88]
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7 0
3 years ago
A small remote-control car with a mass of 1.65 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow
Reil [10]

Answer:

N=63.69N

Explanation:

The normal force exerted on the car by the walls of the cylinder at the bottom of the vertical circle will be such that when substracted to the weight it must give the centripetal force, since at that point on the vertical F_{cp}=N-W=N-mg

We also know that the equation for the centripetal force is:

F_{cp}=ma_{cp}=\frac{mv^2}{r}

Mixing both equations we get:

N-mg=\frac{mv^2}{r}

N=mg+\frac{mv^2}{r}

Which for our values means:

N=(1.65Kg)(9.8m/s^2)+\frac{(1.65Kg)(12m/s)^2}{(5m)}=63.69N

8 0
3 years ago
A velocity vs. time graph is shown.
Scilla [17]

Answer:

Acceleration, a=4\ m/s^2

Explanation:

We are given with a velocity-time graph of an object that starts from origin. y axis is velocity and x-axis is time.

The slope of velocity-time graph gives acceleration of an object. Taking the slope of this graph.

m=\dfrac{v}{t}\\\\m=\dfrac{20\ m/s}{5 s}\\\\m=\text{acceleration}=4\ m/s^2

So, the acceleration of the object is 4\ m/s^2.

8 0
4 years ago
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