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3 years ago

A machine does 1500 joules of work in 30 second What is the power of the machine

1 answer:

8 0

This is simple as power in watts is equal to joules per second so we can do 1500 joules divided by 30 seconds which equals 50 watts

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Birds sitting on a single power line don't get shocked. But if they were to place one foot on each of two lines, ___________ wou

Natalka [10]

Current would flow between them and they would receive a terrible shock.

5 0

2 years ago

Read 2 more answers

a linear function has the same y-intercept as x + 4y equals 16 and it's graph contains the point (4,5). Find the slope of the li

navik [9.2K]

Answer: $\bold{\text{Slope (m)}=\dfrac{1}{4}}$

<u>Explanation:</u>

A linear equation is of the form: y = mx + b where

m is the slope
b is the y-intercept (where it crosses the y-axis)

x + 4y = 16

4y = -x + 16

$y = -\dfrac{1}{4}x+\dfrac{16}{4}$

$y=-\dfrac{1}{4}x+4$

The y-intercept (b) = 4

Next, find the slope given point (4, 5) and b = 4

$y=mx+b\\\\5=m(4)+4\\\\1=4m\\\\\dfrac{1}{4}=m\\\\\\\\\large\boxed{Slope (m)=\dfrac{1}{4}}$

6 0

3 years ago

The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the

Margaret [11]

Answer:

$v=2019.09\ m.s^{-1}$

Explanation:

Given:

time taken by the sun to complete one revolution, $t=601.2\ hr$

radial distance of the sunspot, $r=6.955\times 10^8\ m.s^{-1}$

<u>Therefore, angular speed of rotation of sun:</u>

$\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}$

<u>Now the tangential velocity of the sunspot can be given by:</u>

$v=r.\omega$

$v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}$

$v=2019.09\ m.s^{-1}$

4 0

2 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0

1 year ago

HELP ME please

allochka39001 [22]

Answer:

Ruko zara kuch Time dedo na please

3 0

2 years ago