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3 years ago

What is the voltage of a motor that draws a current of 2 a and produces 240 w of power?

1 answer:

8 0

Here Power = Voltage * Current

So, Voltage = Power/Current

Put the values,

V = 240/2

V = 120 V

In short, Your Final Answer would be: 120 Volts

Hope this helps!

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21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe

DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, $F_f$

Since the two forces are in opposite direction, the equation of motion is

$F-F_f = ma$

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

$F - F_f = 0\\F_f = F$

which means that the magnitude of the force of friction is also equal to 40 N.

Learn more about force of friction here:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

3 0

3 years ago

A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe

Liula [17]

Answer:

a) v₁fin = 3.7059 m/s (→)

b) v₂fin = 1.0588 m/s (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s ⇒ Kin = 0 J

v₁fin = ?

hin = L = 0.7 m

hfin = 0 m ⇒ Ufin = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*hin = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*hin) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s (→)

b) Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒ v₂fin = 1.0588 m/s (→)

7 0

3 years ago

Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f

Step2247 [10]

The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.

8 0

3 years ago

What is the acceleration of a race car if it has a mass of 1200kg and is moving with an engine force of 400N

meriva

Answer:

0.34 m/s^2

Explanation:

force=mass × acceleration

400 =1200 × acceleration

acceleration=400/1200

=0.34 m/s^2

5 0

2 years ago

How does changing the potential difference in a circuit affect the current and the resistance?

Effectus [21]

Answer:

The current will be increased and also for the resistance.

Explanation:

The analysis of a direct current circuit can give us the explanation we need. Using the ohm law, which tells us that the voltage is equal to the product of the current by the resistance we have:

$V=I*R\\where\\V= voltage [V]\\I= amperes [amp]\\R=resistance [ohm]\\$

The voltage is equal to the potential difference therefore we will have these expressions:

$I=\frac{V}{R} \\R= \frac{V}{I}$

If we increase the potential differential or circuit voltage, the current will also increase and so does the resistance by increasing the voltage. If we put numerical values in the equation given before, we can confirm this fact.

4 0

3 years ago