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Arada [10]
3 years ago
14

A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sle

d attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Physics
1 answer:
ycow [4]3 years ago
4 0

Answer:

\mu=0.185

Explanation:

From the question we are told that:

Mass m=17kg

Force F=33N

Velocity v=1.6m/s

Distance d= 9.8m

Generally the equation for Work done is mathematically given by

 W=\triangle K.E+\triangle P.E

Where

 \triangle K.E=(F-F_f)*2

 F_f=F+\frac{\triangle K.E}{d}

 F_f=33+\frac{0.5*17*1.6^2}{9.8}

 F_f=30.8N

Since

 f = \mu*m*g

 \mu= 30.8/(m*g)

 \mu= 30.8/(17*9.81)

 \mu=0.185

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slava [35]

Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?

Answer:

The force = 38.4 N

Explanation:

From coulombs law,

F = kq₁q₂/r² ............................ Equation 1

Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.

When, F = 6.4 N, r = d m.

6.4 = kq₁q₂/d²......................... Equation 1

When q₁' = 2q₁, q₂' = 3q₂, r = d cm

F = k(2q₁)(3q₂)/d²

F = 6kq₁q₂/d².......................... Equation 2

Dividing Equation 1 by equation 2

6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)

6.4/F = 1/6

F = 6.4×6

F = 38.4 N.

Thus the force = 38.4 N

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A 20 Ohm resistance is connected
frutty [35]

Answer: 225 V

Explanation:

<u>Given:</u>

Secondary voltage,V2 = 150v

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Supply current, ie, I1 = 5A

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\text{Again taking,}\frac{N_{1}}{N_{2}}=\frac{V_{2}}{V_{1}}$ \\So, \\$V_{1}=\frac{N_{1}}{N_{2}} V_{2}\\$ inserting all values, we get, $V_{1}=\frac{7.5}{5} * 150=225$\\Therefore, \fbox{Primary voltage, {$V_{1}=225 \mathrm{~V}$}}

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A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .
sp2606 [1]
How much work in J does the string do on the boy if the boy stands still? 

<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>

<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>

<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>

<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>

<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
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