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3 years ago

5

When the forces applied to a body are balanced, the body is said to be in ___________________________________

1 answer:

mafiozo [28]3 years ago

5 0

the body is said to be in equilibrium

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A stationary block resting on the ground is pulled up with a tension force of 100N, but does not leave the ground.

tekilochka [14]

Answer:

The normal force the ground exerts on the block, F = -300 N

Explanation:

Given data,

The block pulled up with a tension force, T = 100 N

The weight of the block, W = 300 N

The weight of the block is due to the force of attraction of gravitation.

The surface exerts a force that is equal and opposite to the force acting on the block due to gravitation.

The weight of the block,

W = mg

300 N

The normal force the ground exerts on the block,

F = - mg

= - 300 N

Hence, the normal force the ground exerts on the block, F = -300 N

6 0

3 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

<em>a) How many parsecs are there in one astronomical unit? </em>

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

<em>(b) How many meters are in a parsec? </em>

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

<em>(d) How many astronomical units in a light-year?</em>

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

<em>(e) How many light-years in a parsec?</em>

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

4 years ago

Electromagmetic radiation comes in different forms which pair does not appear next to one an other in the electromagmetic spectr

vivado [14]

Answer:

A) microwaves and ultraviolet

Explanation:

this is the spectrum in order:

radio waves

microwaves

infra red

visible light

ultraviolet

X rays

gamma rays

6 0

3 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

o-na [289]

Answer:

Current = 10 Amperes.

Explanation:

Given the following dat;

Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C

Time = 1 hour to seconds = 60*60 = 3600 seconds

To find the current;

Quantity of charge = current * time

Substituting in the equation

36000 = current * 3600

Current = 36000/3600

Current = 10 Amperes.

6 0

3 years ago

Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?

Ainat [17]

PV = 400 x 0.08 = 32 J

Hope this helps

7 0

3 years ago