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2 years ago

9. If 100 J of work is done in 20 second the power is ..

Physics

2 answers:

Bumek [7]2 years ago

7 0

5 watts fjfjfjdjdjdjjdjdjdjfjd

Mamont248 [21]2 years ago

6 0

Answer is 5 watts.

Using the equation:
w=pt

W = work done (J)
P = power (Watts)
T = Time (s)

Substitute the values into the equation.

100 = 20p

Power = 5 watts

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Travka [436]

Answer:

d I believe so

Explanation:

d yes?????

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2 years ago

A 7.0-μC point charge and a point charge are initially extremely far apart. How much work does it take to bring the point charge

vazorg [7]

Answer:1.008 ×10^-14/rJ

Where r is the distance from.which the charge was moved through.

Explanation:

From coloumbs law

Work done =KQq/r

Where K=9×10^9

Q=7×10^-6C

q=e=1.6×10^-19C

Micro is 10^-6

W=9×10^9×7×10^-6×1.6×10^-19/r=100.8×10^-16/r=1.008×10^-14/rJ

r represent the distance through which the force was used to moved the charge through.

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2 years ago

Read 2 more answers

Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

kifflom [539]

Answer:

2 /s north

Explanation:

Given that,

Velocity due North is 8 m/s and due south is 6 m/s

We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

$v=8+(-6)\\\\v=2\ m/s$

It is positive. So, it is in North direction.

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3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu

galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

$F=I \times B \times L \times sin(\theta)$