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melisa1 [442]
3 years ago
15

A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2

.5m on the floor. calculate the coefficient of sliding friction [g = 10.0 ms per second per second]
Physics
1 answer:
erastova [34]3 years ago
5 0
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps




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Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

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peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V

The rms voltage (in V) measured across the secondary coil is calculated as;

V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

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brainly.com/question/6536722

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