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3 years ago

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The force that contributed to the formation of planets, determines the motion of bodies in the solar system, and pulls objects t

o the center of Earth is called

1 answer:

kykrilka [37]3 years ago

6 0

Answer:

The force of gravity

Explanation:

Gravity was studied, by early scientists such as Copernicus and others, Galileo was the first to ensure that planets moved according to a physical equation that depended on a force that caused celestial bodies to move and interact with each other. But years later Newton based on studies conducted deciphering what Galileo assumed, he was able to find the equation of the force of gravity in any body in the universe. This equation depends on the masses of the two interacting bodies, the distance between them and a constant, which I call universal gravitation constant.

$F_{g}=G*\frac{m_{1}*m_{2}}{r^2}$

Fg = gravity force [N]

G = universal gravitation constant = 6.67*10^(-11) [N*m^2/kg^2]

m1 = mass of the 1st body [kg]

m2 = mass of the 2nd body [kg]

r = distance between the bodies [meters]

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Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.

finlep [7]

Answer:

$\tau=(-32k)\ N-m$

$\theta=116.55^{\circ}$

Explanation:

Given that.

Force acting on the particle, $F=(-6i + 2.0j)\ N$

Position of the particle, $r=(4i+4j)\ m$

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

$\tau=F\times r$

$\tau=(-6i + 2.0j)\times (4i+4j)$

$\tau=\begin{pmatrix}0&0&-32\end{pmatrix}$

$\tau=(-32k)\ N-m$

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, $|r|=\sqrt{4^2+4^2}=5.65\ m$

Magnitude of F, $|F|=\sqrt{(-6)^2+2^2}=6.324\ m$

Using dot product formula,

$F{\circ}\ r=|F|.|r|\ cos\theta$

$cos\theta=\dfrac{F{\circ} r}{|F|.|r|}$

$cos\theta=\dfrac{-24+8}{6.324\times 5.65}$

$\theta=116.55^{\circ}$

Therefore, this is the required solution.

6 0

3 years ago

Read 2 more answers

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the <em>book is</em>

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

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2 years ago

Use what you know about reflection and absorption of light to complete the sentence.

Blababa [14]

A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.

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3 years ago

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19. State any 3 applications of capillary action (3mk)

zlopas [31]

Explanation:

1. Movement of water, food and mineral salts in plants

2. Absorption of water by towels when wiping our bodies

3. It is used to absorb ink using a blotting paper or tissue

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3 years ago

A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate

yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

$W=Pm\DeltaV$

$W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}$

$W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})$

Where,R = gas constant

T = temperature

P = pressure

$P_{atm}$ =Atmosphere pressure

m = mass

Put the value into the formula

$W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)$

$W=213\ kJ$

Hence, The work done by the steam is 213 kJ.

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3 years ago