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tankabanditka [31]
3 years ago
13

A laptop computer draws 5 A of current with a voltage of 19 V. Which is the power used by the computer?

Physics
2 answers:
Sav [38]3 years ago
6 0
There are several information's of importance already given in the question. Based on those information's the answer can be deduced.
Volage = 19 V
Current = 5 A
Power = Voltage x Current
           = 19 x 5
           = 95 Watt
From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the second option or 95 W.
nataly862011 [7]3 years ago
6 0

Answer : The power used by the computer is, 95 W

Solution : Given,

Current = 5 Ampere

Voltage = 19 Volt

Power : It is defined as the product of the voltage and the current in the circuit.

Formula used :

P=V\times I

where,

P = power

V = voltage

I = current

Now put all the given values in the above formula, we get the power used by the computer.

P=V\times I=(19V)\times (5A)=95watt=95W

Therefore, the power used by the computer is, 95 W

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A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

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F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

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6 0
3 years ago
A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete
lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

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(a)

If friction neglected,

⇒ F_x=mgSin \theta=ma

⇒ a_x=gSin \theta

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(b)

If friction present,

⇒ F_x=mgCos \theta

⇒ F_x=mgSin \theta-\mu_r mgCos \theta

⇒ a=gSin \theta-\mu_rgCos \theta

      =9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}

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SVETLANKA909090 [29]

Answer:

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