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2 years ago

Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

Physics

1 answer:

liubo4ka [24]2 years ago

7 0

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

$a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N$

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

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2 years ago

A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)

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Answer:

205 V

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Explanation:

L = Inductance in Henries, (H) = 0.500 H

resistor is of 93 Ω so R = 93 Ω

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$V_{L} = - IwLsin(wt)$

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

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I = 0.044 A

Now

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Deriving formula for voltage across the resistor

The derivative of sin is cos

V $_{R}$ = V $_{R}$ cos (wt)

Putting V $_{R}$ = 4.092 V and w = 500 rad/s

V $_{R}$ = V $_{R}$ cos (wt)

= (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V $_{R}$ = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

= (4.092 V) (cos (500 rads/s)(0.00209))

= (4.092 V) (cos(1.045))

= (4.092 V)(0.501902)

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