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omeli [17]
3 years ago
15

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

40 kg, calculate the mass of another body. (And:119.9 kg)​
Physics
1 answer:
cluponka [151]3 years ago
3 0

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

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adelina 88 [10]

The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.

<h3 /><h3>What is mass?</h3>

Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)

To calculate the mass the engine burns each seconds, we use the formula below.

Formual:

  • M = T/v............. Equation

Where:

  • M = Mass per seconds of the rocket
  • T = Thrust
  • v = Velocity

From the question,

Given:

  • T = 7.66×10⁵ N
  • v = 3.05×10³ m/s

Substitute these values into equation 1

  • M = (7.66×10⁵)/(3.05×10³)
  • M = 2.5×10² kg/s

Hence, the mass of fuel burned in each second is 2.5×10² kg/s.

Learn more about mass here: brainly.com/question/25121535

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8 0
1 year ago
The rest deltoid row is a back exercise true or false
bixtya [17]
False because your deltoids are in your shoulders not your back
3 0
3 years ago
What was an ill effect of the industrial revolution?
Digiron [165]

Answer:

answer is option (c) child labour

6 0
3 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s
PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0
2 years ago
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