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2 years ago

Please help I need help I can’t fail .

Physics

1 answer:

Evgen [1.6K]2 years ago

6 0

Answer:

Explanation:

Gas particles move fastest.

Plasma lacks fixed volume and shape.

Liquid lacks fixed shape, but has fixed volume.

Solid has both fixed volume and shape.

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If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

Bas_tet [7]

Answer:

W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

F = G m₁ M / r²

W = ∫ F. dr

W = G m₁ M ∫ dr / r²

we integrate

W = G m₁ M (-1 / r)

We evaluate between the limits, lower r = R_ Moon and r = ∞

W = -G m₁ M (1 /∞ - 1 / R_moon)

W = G m1 M / r_moon

Body weight is

W = mg

m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

m = 1000/32 = 165 / g_moon

g_moom = 165 32/1000

.g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system

W_capsule = 1000 pound (1 kg / 2.20 pounds)

W_capsule = 454 N

W = m_capsule g

m_capsule = W / g

m = 454 /9.8

m_capsule = 46,327 kg

Let's calculate

W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶

W = 1,307 10⁶ J

7 0

2 years ago

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?

joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

3 0

2 years ago

An woman whose weight is 804 N stands on a long horizontal plank of wood 1.55 m from one end. The plank is uniform and is suppor

meriva

Answer:

Explanation:

There will be reaction force by each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman

Taking torque of all forces about the end far away from the woman

Torque by reaction force = R₁ x 5.5

= 5.5 R₁ upwards

Torque by weight of woman in opposite direction , downwards

= - 804 x ( 5.5 - 1.55 )

= - 3175.8

Torque by weight of the plank in opposite direction , downwards .

= - 27 x 5.5 / 2

= - 74.25

Torque by R₂ will be zero as it passes through the point about which torque is being taken .

Total torque

= 5.5 R₁ - - 3175.8 - - 74.25 = 0 ( For equilibrium )

5.5 R₁ = 3250

R₁ = 590.9 N .

6 0

2 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

2 years ago

Svetlanka [38]

Answer:

D . A mass of 5 kilograms lifted 5 meters in 10 second

4 0

2 years ago