Answer:
First let's find the work.
Work = Force*Displacement
= 20*5
= 100 Joules
Power = Work/Time
= 100/4
= 25 Watts
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Answer:
Cardiac Arrest, burns, and nerve damage.
Explanation:
Basically, the main risk is cardiac arrest, caused by the electric current interfering with the normal operation of the heart muscle. Other possible damages are burns due to the electric energy vaporizing the water inside the cells, and nerve damage caused by excessive current through the nerves.
Hi there!
Recall the equation for spring potential energy:
k = Spring constant (N/m)
x = extension of spring from equilibrium (m)
PE = Potential Energy (J)
Plug in the given values:
I'm pretty sure you can find it out by using a speed monitor and compass or just observing it
if there any answer choices tell me.
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