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3 years ago

7. A type of non-renewable energy would be: O wind O petroleum O water solar

Physics

1 answer:

matrenka [14]3 years ago

4 0

Answer:

petroleum

Explanation:

Petroleum is a gas that we use that just goes into the air whilst water can be recycled and purified, wind never ever goes away, and solar power can be converted into energy.

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In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev

myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

$10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB$

where;

$I_o$ = initial intensity

$log \bigg(\dfrac{I}{I_o} \bigg) = 1.3$

$\dfrac{I}{I_o}= 10^{1.3 }$

I = 19.95 $I_o$

I ≅ 20 pig callers

6 0

3 years ago

Principal of simple machine?

sergij07 [2.7K]

Answer:

if there is no friction in a simple machine, work output and work input are found equal in that machine

Explanation:

3 0

3 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

levacccp [35]

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

5 0

3 years ago

I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving

snow_lady [41]

Answer:

$9.7\times 10^{-4}\ rad/s$

Explanation:

Given:

$m=500 kg\\F=3000 N$

$Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\ know\ that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}$

= $\frac{6}{6371 \times 10^3\ m}$

= $9.7\times 10^{-4}\ rad/s$

7 0

3 years ago

NEED THIS DONE ASAP HELP

ExtremeBDS [4]

Answer: Exercise Physiology

Explanation:

4 0

3 years ago