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4 years ago

Define s.i unit of force explanation

Physics

1 answer:

kvv77 [185]4 years ago

3 0

<h3><u>Answer</u> :</h3>

◈ As per newton's second law of motion, Force is defined as the product of mass and acceleration $.$

Mathematically,

$\dag\:\boxed{\bf{\purple{F=ma}}}$

Unit of mass : kg

Unit of acceleration : m/s²

Therefore,

Unit of force ➠ <u>kg m/s²</u>

SI unit : <u>N (newton)</u> or <u>kg m/s²</u>

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Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 170 V. An electron ent

notsponge [240]

Answer:

Explanation:

Electric field between plates

V / d

= 170 / ( 2 x 10⁻² )

= 8500 N/C

Force on electron in this field

= 8500 x 1.6 x 10⁻¹⁹

= 13600 x 10⁻¹⁹ N

Acceleration

= 13600 x 10⁻¹⁹ / 9.1 x 10⁻³¹

a = 1494.5 x 10¹² m /s²

s = .1 x 10⁻² m

v² = u² + 2as

= (2.9x 10⁵)²+ 2 x 1494.5 x 10¹² x .1 x 10⁻²

= 8.41 x 10¹⁰ + 299 x 10¹⁰

= (8.41 + 299 ) x 10¹⁰

v = 17.53 x 10⁵ m /s

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3 years ago

The part of the cerebrum that controls voluntary movements is located in the _______ lobe. A. frontal B. parietal C. temporal D.

uranmaximum [27]

The part of the cerebrum that controls Voluntary movements is located in the Frontal lobe.

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3 years ago

Read 2 more answers

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is: