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Marta_Voda [28]
4 years ago
8

Recently, some information about a distant planet was learned. It has a radius of 6000000 meters, and the density of the atmosph

ere as a function of the height h (in meters) above the surface of the planet is given by δ(h)=3h+6000000 kilograms per cubic meter. Calculate the mass of the portion of the atmosphere from h=0 to h=57.
Physics
1 answer:
Oksana_A [137]4 years ago
7 0

Answer:

M = 9.8 ×10²²kg

Explanation:

This question involves the mass, density and volume relationship. The density of the atmosphere varies with the height above the surface of the planet. Given the density function δ(h)=3h+6000000. We can then calculate the value of the densities at the two given altitudes (height above the planet surface).

We will use the relationship between the mass, volume and density.

M = ρ × V

At h = 0m (the surface of the planet)

The radius of the planet = 6×10⁶m

ρ = 3h+6000000 = 3×0 + 6000000

= 6000000 = 6×10⁶ kg/m³

V = 4/3× r³ (volume of a sphere)

= 4/3× (6×10⁶)³ = 2.88 ×10²⁰ m³

M = 6×10⁶ × 2.88 ×10²⁰ = 1.728 ×10²⁷

At a height of h = 57m

r = 6000000 + 57 = 6000057m

V = 4/3 × (6000057)³ = 2.880082×10²⁰

ρ = 3h+6000000 = 3×57 + 6000000

= 6000171 kg/m³

M = 6000171 × 2.880082 × 10²⁰

= 1.728098 × 10²⁷

The mass of the atmosphere is the difference between the masses at the different altitudes.

So the mass of the atmosphere

= 1.728098 × 10²⁷ – 1.728000 × 10²⁷

= 9.8 ×10²² kg.

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leva [86]

Answer:

The work done required on the coin during the displacement is 21.75 w.

Explanation:

Given that,

A coin slides over a friction-less plane i.e friction force = 0.

The co-ordinate of the given point is (1.40 m, 7.20 m).

The position vector of the given point is represented by  1.40 \hat i+7.20 \hat j.

The displacement of the coin is

\vec d=1.40 \hat i+7.20 \hat j

The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.

Then x component of the force = 4.50 cos128°

The y component of the force = 4.50 sin128°

Then the position vector of the force is

\vec F=(4.50 cos 128^\circ)\hat i+(4.50 sin 128^\circ)\hat j

   =-2.77 \hat i+3.56 \hat j

We know that,

work done is a scalar product of force and displacement.

W=\vec F.\vec d

    =(-2.77 \hat i+3.56 \hat j).(1.40 \hat i+7.20 \hat j)

    =(-2.77×1.40+ 3.56×7.20) w

    =21.75 w

The work done required on the coin during the displacement is 21.75 w.

6 0
3 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

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Using hookes law find the distance a spring with am elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it
siniylev [52]
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4 years ago
A beam of light passes from air into a cube of zirconium if the angle of incidence is 40”what is the angle of refraction ?
Colt1911 [192]

Answer:

50 degrees

Explanation:

Usually in physics in solving such questions refractive index of air is taken as 1.00 (n=1.00) by assuming that air is a vacuum. Hence, the light ray passes from water to air at an angle of incident of 40 degrees, it means that it has angle of 50 degrees to the normal line.

5 0
4 years ago
A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w
nignag [31]

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

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  • v is velocity
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265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

  • vyf is the final vertical velocity
  • vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

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  • vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

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vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1  

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