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3 years ago

Describe three ways you can conserve energy in your own home

Physics

2 answers:

prohojiy [21]3 years ago

7 0

You can turn off the air conditioning
you can keep all electronics off
you can turn off all running water

Mandarinka [93]3 years ago

4 0

Turn lights off, unplug electronics, and use solar energy

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A worker pushes horizontally on a 36.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the

Yuliya22 [10]

Answer:

value of fs,max under the circumstances is

f = 42.56 N

Explanation:

The formula is µ = f / N, where µ is the coefficient of friction, f is the amount of force that resists motion, and N is the normal force. Normal force is the force at which one surface is being pushed into another.

So f = µN

f = 0.380 * 112

f = 42.56 N

5 0

3 years ago

Read 2 more answers

Which state of matter has a definite volume but no definite shape

a_sh-v [17]

The answer is liquid.

6 0

3 years ago

In which electric circuit would the voltmeter read 10 volts ?

ki77a [65]

Given that,

Voltage = 10 volt

Suppose, The three resistance is connected in parallel and each resistance is 12 Ω. find the current in the electric circuit.

We need to calculate the equivalent resistance

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}$

$\dfrac{1}{R}=\dfrac{1}{4}$

$R=4\ \Omega$

We need to calculate the current in the circuit

Using ohm's law

$V=IR$

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

Where, V = voltage

R = resistance

Put the value into the formula

$I=\dfrac{10}{4}$

$I=2.5\ A$

Hence, The current in the circuit is 2.5 A

4 0

3 years ago

a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem

VikaD [51]

W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

6 0

3 years ago

Read 2 more answers

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

$m = 25 * 12.5$

$m \approx 310$

6 0

3 years ago