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3 years ago

A backpack weighs 8.2 Newtons and has a mass of 5 kilograms on the moon what is the strength of gravity on the moon

Physics

1 answer:

lions [1.4K]3 years ago

7 0

i dont get it so much but

The weight of the bag pack is 8.2 N. g = 1.64 m/s2. Hence, the acceleration due to gravity on moon is 1.64 m/s2. sooo? is it right

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A wave that can travel with or without a medium is called a(n)

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A bus travelling 30.0 km/h east has a constant increase in speed of 1.5 m/s. What is its velocity 6.8s later

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3 years ago

Find the mass of a body if the acceleration the body used to move is given as 5 m/s-2 and the force of the body is 30N take g=10

Aleonysh [2.5K]

The mass of a body if the acceleration the body used to move is given as 5 m/s-2 will be 3 kg.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Force,F = 30 N

Mass,m = kg

Acceleration,a = 5 m/s²

The force is found as;

F=ma

30 N =m kg × 5 m/s²

m=3 kg

Hence the mass of a body will be 3 kg.

To learn more about the force refer to the link;

brainly.com/question/26115859#SPJ1

#SPJ1

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2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago