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3 years ago

What is the energy equivalent of an object with a mass of 25 kg?

Physics

1 answer:

Amanda [17]3 years ago

3 0

Answer:

$2.25\cdot 10^{18} J$

Explanation:

The energy equivalent of the mass of an object can be calculated by using the Einstein's equation:

$E=mc^2$

where

E is the energy

m is the rest mass of the object

$c=3\cdot 10^8 m/s$ is the speed of light

For this object

m = 25 kg

So its equivalent energy is:

$E=(25)(3\cdot 10^8)^2=2.25\cdot 10^{18} J$

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"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat

vagabundo [1.1K]

Answer:

(a) g = 8.82158145 $m/s^2$ .

(b) 7699.990192m/s.

Explanation:

(a) Strength of gravitational field 'g' by definition is

$g = \frac{M_{(earth)} }{r^2} G$ , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145 $m/s^2$ .

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

$a_{centripetal}=\frac{V^2}{r} =g$ here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c) S = vT, here T is time period or time required to complete one full revolution.

S = earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

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3 years ago

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3 years ago

If we increase the force applied to an object and all other factors remain the same that amount of work will

worty [1.4K]

Hello there.

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</span><span>C. Increase
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3 years ago

Read 2 more answers

Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

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3 years ago

you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

Sergio039 [100]

Some people view bacteria specimens with a 100x objective lens in order to see the smallest details.
Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
And still others may use a 40x objective lens to gain maximum resolution when viewing images of thick samples.

It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>

Bacteria must, of course, be viewed at the maximum magnification and resolution possible because to their small size.
Due to optical restrictions, this is approximately 1000x in a light microscope.
To improve resolution, the oil immersion method is performed. This calls for a unique 100x objective.

To learn more about bacterial specimen, visit:

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2 years ago