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2 years ago

A wire from the power supply is carrying 120w of power and 24a of current. which color(s) of cable is the wire

2 answers:

8 0

There is no picture given so I can't be really sure what color of the cable you're referring to. However, the only relationship I can think of when the power and the current is given would be: P=IV or P = I²R, where P is power, I is current, V is voltage and R is resistance. Solving both equations:

120 W = (24 A)(Voltage)
Voltage = 5 V

120 W = (24 A)²(R)
R = 0.2083 Ω

So, i think the cable would have specification of 5 Volts and 0.2083 ohms.

boyakko [2]2 years ago

8 0

Answer and explanation;

Power = voltage × current;

P = V/I

Thus;

V = P/I

= 120W/24A

= 5V

The voltage is therefore; 5 volts

The wire color is either Red or White.

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

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Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, $m=5.0$ kg

Force acting towards east, $F=27$ N

Frictional force acting towards west, $f=17$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards east is given as:

$F_{net}=F-f=27-17=10\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Therefore, the magnitude of the acceleration of the box is 2 m/s².

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2 years ago

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Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.

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Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, $q_1=6\ \mu C=6\times 10^{-6}\ C$

Charge 2, $q_2=2\ \mu C=2\times 10^{-6}\ C$

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

$F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N$

So, the force applied between charges is 10.788 N.

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2 years ago

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A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction

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Answer:

F=UR

52.9=U*46.6

U=52.9/46.6

U=1.135

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A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie

cluponka [151]

Answer:

a) v₃ = 19.54 km, b) 70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

cos 45 = x₁ / V₁

sin 45 = y₁ / V₁

x₁ = v₁ cos 45

y₁ = v₁ sin 45

x₁ = 26 cos 45

y₁ = 26 sin 45

x₁ = 18.38 km

y₁ = 18.38 km

Vector 2 moves 45 km north

y₂ = 45 km

Unknown 3 vector

x3 =?

y3 =?

Vector Resulting 70 km north of the starting point

R_y = 70 km

we make the sum on each axis

X axis

Rₓ = x₁ + x₃

x₃ = Rₓ -x₁

x₃ = 0 - 18.38

x₃ = -18.38 km

Y Axis

R_y = y₁ + y₂ + y₃

y₃ = R_y - y₁ -y₂

y₃ = 70 -18.38 - 45

y₃ = 6.62 km

the vector of the third leg of the journey is

v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

v₃ = √ (18.38² + 6.62²)

v₃ = 19.54 km

to find the angle let's use trigonometry

tan θ = y₃ / x₃

θ = tan⁻¹ (y₃ / x₃)

θ = tan⁻¹ (6.62 / (- 18.38))

θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

θ’= 180 -19.8

θ’= 160.19º

I mean the address is

θ’’ = 90-19.8

θ = 70.2º

70.2º north-west

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2 years ago

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

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