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3 years ago

How much kinetic energy does a person of mass 50kg running at 3m/s have? Provide

Physics

1 answer:

ddd [48]3 years ago

4 0

Answer:

0.225 KJ kinetic energy is used for a person of mass 50 Kg running at speed of 3 m/s.

Explanation:

Given:

mass = 50 kg

velocity = v = 3 m/s

To Find :

Kinetic energy = ?

Solution:

Kinetic energy of an object is the energy that it possesses due to its motion.

The formula for kinetic energy is given as

$\textrm{kinetic energy} = \frac{1}{2}\times mass\times velocity^{2}\\= \frac{1}{2}\times m\times v^{2}\\= \frac{1}{2}\times 50\times 3^{2}\\= 25\times 9\\= 225\ Joules.$

But we required kinetic energy in KiloJoules i.e KJ

1 K = 1000 = $10^{3}$

$\textrm{kinetic energy} = 225\ J\\= 0.225\times 1000\\= 0.225 \times 10^{3}\\= 0.225\ KJ$

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A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

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3 years ago

What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?

babunello [35]

Answer:

m³/(kg⋅s²)

Explanation:

Hello.

In this case, since the involved formula is:

$F=G*\frac{m_1m_2}{r^2}$

By writing a dimensional analysis with the proper algebra handling, we obtain:

$N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}$

Thus, answer is:

m³/(kg⋅s²)

Note that the [=] is used to indicate the units of G.

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2 years ago

Why do you think bacteria can live in your digestive system without making you sick?

Vikentia [17]

Bacteria in your stomach is actually good for you because it helps destroy harmful bacteria by picking at undigested leftovers and micromanaging calories.

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3 years ago

Read 2 more answers

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is acc

levacccp [35]

The formula for resonant frequency is:

$f_0=\frac{1}{2\pi\sqrt{LC} }$

Given information:

$f_{0\text{,small}}=500 \text{ kHz}\\f_{0\text{,large}}=1650 \text{ kHz}\\L=3.83\text{ } \mu \text{H}$

Plug in the given values to find one value of capacitance:

$500 \text{ kHz}=\frac{1}{2\pi\sqrt{C(3.83\text{ } \mu \text{H})} }\\C=2.645*10^{-8} \text{ F}=26.45 \text{ nF}$

Plug in the given values to find the other value of capacitance:

$1650 \text{ kHz}=\frac{1}{2\pi\sqrt{C(3.83\text{ } \mu \text{H})} }\\C=2.429*10^{-8} \text{ F}=2.429 \text{ nF}$

This gives a range of 2.429 nF to 26.45 nF.

With significant figures taken into account, the range of capacitance is 2.43 nF to 30 nF.

6 0

2 years ago

Read 2 more answers

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

Kipish [7]

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

$\beta = 10Log(\frac{I}{I_o} )\\\\$

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

$120 = 10Log(\frac{I}{10^{-12}} )\\\\12 = Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12} * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2$

The intensity of sound of a whisper;

$20 = 10Log(\frac{I}{10^{-12}} )\\\\2 = Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2} * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\$

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

$\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}$

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

3 0

3 years ago