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Marina86 [1]
4 years ago
12

At a constant velocity, you are pulling a 10kg object with. Force P. How much work is done by P to move 87cm? Assume kinetic fri

ctional coefficient = 0.26.
Physics
1 answer:
lara [203]4 years ago
3 0
Constant velocity means............
applied force=frictional force=p=kinetic friction coefficient*mg=0.26*10*9.8=?
now we know ....  work=force*distance=0.26*10*9.8*0.87joule
plz feel free to ask if u hv any confusion!!
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Which was not a part of faradays law regarding the induced voltage in a coil?
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Your answer would be C. Hope this helps!!
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3 years ago
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Which one of the following formulas is used to find the voltage of a circuit? 
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D , since Voltage is one joule per coulomb
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After t hours a freight train is s(t) = 18t2 − 2t3 miles due north of its starting point (for 0 ≤ t ≤ 9). (a) Find its velocity
BartSMP [9]

Answer:

Explanation:

Given the equation modelled by the height of the train given as:

s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9

a) Velocity is the rate of change of displacement.

Velocity = dS(t)/dt

V = dS(t)/dt = 36t - 6t² miles

Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.

V = 36(3) -6(3)²

V= 108 - 72

Velocity = 36mi/hr

b) for Velocity at time = 7hrs

V(7) = 36(7) - 6(7)²

V(7) = 252 - 294

V(7) = -42mi/hr

The velocity at t = 7hrs is -42mi/hr

c) Acceleration is the rate of change of velocity.

a(t) = dV(t)/dt

Given v(t) = 36t - 6t²

a(t) = 36 - 12t

Acceleration at t=1 is given as:

a(1) = 36 -12(1)

a(1) = 24mi/hr²

4 0
3 years ago
A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi
Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N  (b)  Block mass of 4 Kg  [tex]T'-mg sin \theta=0 hence T'=mg sin \theta where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than mg sin\theta , then the box moves down the slope  

(c)  

Acceleration a= \frac {forward   force-backward   force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}  

a= \frac {22.51-19.62}{2+4}=0.48

When moving, the box will exert force T"= mgsin \theta + ma  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0
4 years ago
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1) A person slides a box down a ramp. The box starts from rest 2m above the lowest
wel

Answer:

.98m

Explanation:

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