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2 years ago

A man has a mass of 110kg. What is his weight?

Physics

1 answer:

Stells [14]2 years ago

5 0

His weight depends on where he is, because

Weight = (mass) x (gravity in the place where the mass is) .

For example:

-- If this man is on Mars, his weight is (110 kg) x (3.7 m/s²) = 408 Newtons

-- If he is on the Moon, his weight is (110 kg) x (1.6 m/s²) = 176 Newtons

-- If he is on Earth, his weight is (110 kg) x (9.8 m/s²) = 1,078 Newtons

-- If he is in a spacecraft coasting from one to another, his weight is zero.

You might be interested in

At what distance will a 80 W lightbulb have the same apparent brightness as a 120 W bulb viewed from a distance of 40 m

liq [111]

Answer:

32.6mm

Explanation:

Using area of a sphere(bulb) = 4πr²

So A is proportional to radius²

So the Energy will be proportional to r²

But 120/80 = 1.5 is the energy factor so

Using

1.5/d² = 1/r²

1.5/40²= 1/r^2

r = √( 40²/ 1.5)

r = 32.6m

4 0

3 years ago

g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo

mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D) Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 , exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0

2 years ago

A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

QveST [7]

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

$W = mg = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2$

b) at either end

$P = \frac{F}{A} = P_o + \rho gh$

$dF = PdA$

$dF = \rho g w \int h dh$

$F = \rho g w \frac{h^2}{2}$

$F = \rho g A \frac{h}{2}$ [A = wh]

$F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}$

F = 294000 N

3 0

3 years ago

The spherical annulus of cometary nuclei surrounding the solar system is called the ___.

Bingel [31]

Answer:

Oort Cloud

Explanation:

Hundreds of millions of comet nuclei forming a spherical region that surrounds the solar system is called Oort cloud.

The Oort cloud revolves at tens of thousands of the distance of the earth from the sun.

Comets are the condensed water and dust particles that orbit around the sun with inward helix leaving behind a tail of vapours and vaporize completely as they get closer to the sun during their revolution around it.

They have a nebulous appearance and extremely elongated orbits.

3 0

3 years ago

. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T

Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

I = F t = Δp

in this case the final velocity is zero

F t = 0 -m v₀

F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

F₁ = -p₀ / 0.1

F₁ = - 10 p₀

barrier collapses

F₂ = -p₀ / 1

let's look for the relationship of the forces

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0

3 years ago