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2 years ago

10

The following are connected in parallel circuit at home EXCEPT:

1 answer:

nikdorinn [45]2 years ago

8 0

Light bulbs, think about parallel circuits as something you plug into, so for example you plug in a tv for it to work, same with a refrigerator and Christmas lights

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Elza [17]

1.cool down
2.activity log
3.specific warm up
4.activities of daily living
5.planned exercise
6.general warm up

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3 years ago

What is the speed of sound in air when the temperature is 20°C?

Anarel [89]

The answer to your question is 343 m/s

5 0

3 years ago

Read 2 more answers

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

a)

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

b)

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

2 years ago

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$

4 0

2 years ago

What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

masya89 [10]

Answer: $0.0058 m/s^{2}$

Explanation:

Centripetal acceleration $a_{C}$ is calculated by the following equation:

$a_{C}=\frac{V^{2}}{r}$

Where:

$V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s$ is the Earth's orbital speed

$r=1.5(10)^{11} m$ is the orbital radius

$a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}$

$a_{C}=0.0058 m/s^{2}$

4 0

3 years ago