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3 years ago

What is a unit of mass?

1 answer:

3 0

Units such as kilograms (about 2.2 pounds)n which are a thousand grams,grams,centigrams (hundredths of grams), and milligrams (thousands of grams).

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Explain the relation between Joule and Erg.

LekaFEV [45]

Answer:

Both erg and Joule are units of Energy in the CGS and S.I system respectively.

Joule and erg can also be written as (kg m2/s2) and (g cm2/s2) respectively.

Explanation:

1 Joule = 107 erg ⇒ 107 kg m2/s2.

8 0

1 year ago

Read 2 more answers

A disturbance sends ripples across water in a tub. These ripples are an example of a rarefaction

Zielflug [23.3K]

The appropriate response is a Surface wave. It is a seismic wave that goes over the surface of the Earth rather than through it. Surface waves as a rule have bigger amplitudes and longer wavelengths than body waves, and they travel more gradually than body waves do. Adore waves and Rayleigh waves are sorts of surface waves.

3 0

3 years ago

How to find aceleration

lions [1.4K]

Answer:

$a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}}$

Explanation:

Formula for AVERAGE acceleration

$a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}}$

$v_{f}$ is Final Velocity

$v_{i}$ is Initial Velocity

$t_{f}$ is Final Time

$t_{i}$ is Initial Time

Most of the time, the initial time will be 0, so you won't have to subtract.

5 0

2 years ago

A myopic (nearsighted) lawyer has a far point of 60.0 cm. What is the diopter value of suitable corrective contacts?

Vlada [557]

Answer:

P = -1.67 dP

Explanation:

given,

myopic lawyer has a far point of = 60.0 cm = 0.6 m

If a person is suffering from myopia then he cannot see the farthest object clearly.

Image of far object does not form on the retina so, the image appear to be blur.

using lens formula

$\dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}$

$-\dfrac{1}{0.6} = \dfrac{1}{f}$

$\dfrac{1}{f} =P$

$P = -\dfrac{1}{0.6}$

P = -1.67 dP

6 0

2 years ago

A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

7 0

3 years ago