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3 years ago

What is the relationship between SI units and CGS unit of work ?

1 answer:

8 0

Work SI unit = joule = N*m= [kg]*[m/s^2] *[m] = kg * m^2/s^2

Work cgs unit = erg = [g][cm/s^2][cm] = g*cm^2 / s^2

Then 1 kg * m^2 / s^2 * [1000 g/kg] * [100cm/m]^2 = 10,000,000 g*cm^2/s^2

The relation is 1 joule = 10,000,000 erg or 1 erg = 10^-7 joule

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Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

vladimir2022 [97]

Answer:

The distance is 300 m.

Explanation:

Given that,

Time = 30 s

Speed = 80 m/s

Distance = 1200 m

Speed of smaller plane = 40 m/s

We need to calculate the acceleration

Using equation of motion

$s= ut+\dfrac{1}{2}at62$

Put the value in the equation

$1200=0+\dfrac{1}{2}\times a\times(30)^2$

$a=\dfrac{2\times1200}{30\times30}$

$a=2.67\ m/s^2$

We need to calculate the distance

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$40^2=0+2\times2.67\times s$

$s=\dfrac{40^2}{2\times2.67}$

$s=299.62\approx 300\ m$

Hence, The distance is 300 m.

3 0

2 years ago

Which describes the eye of a hurricane?

777dan777 [17]

<h2><em><u>B</u></em></h2><h2><em><u>THE EYE OF THE STORM IS CALM</u></em></h2>

7 0

2 years ago

Read 2 more answers

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

Which of the following is the best definition of the term pseudopsychology?

blagie [28]

Can you give us the options…?

3 0

2 years ago

A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.

harina [27]

I can not solve the problem if I do not have the mass.

3 0

2 years ago