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xz_007 [3.2K]
3 years ago
13

Two forces are applied to a 2.0 kg block on a frictionless horizontal surface. F1 = 8.ON is applied to the left while F2 = 3.0 N

is applied to the right. What is the
acceleration of the block?


A.) 2.5 m/s^2 to the left

B.) 1.5 m/s^2 to the right

C.) 4.0 m/s^2 to the left

D.) 2.5 m/s^2 to the right
Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

It is C 5.0 m/s 2 to the right

Explanation:

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harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

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then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

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Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

            Where: m is the object's mass

                         g is the  gravitational acceleration in the surface earth

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w_{uc} = 5292 N

And the Honda Accord's weight is:

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w_{HA} = 14210 N

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