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ExtremeBDS [4]
4 years ago
5

What will an object weigh on the Moon's surface if it weighs 100 N on Earth's surface? (b) How many Earth radii must this same o

bject be from the center of Earth if it is to weigh the same as it does on the Moon?
Physics
1 answer:
olga nikolaevna [1]4 years ago
4 0

Answer:

Part a)

W = 16.7 N

Part b)

r = 2.45 R

Explanation:

Part a)

As we know that acceleration due to gravity on the surface of moon is 1/6 times the gravity on the surface of earth

So the force due to gravity will decrease by the factor of 6

so we will have

W_{moon} = \frac{1}{6}W_{earth}

W_{moon} = \frac{1}{6}(100)

W_{moon} = 16.7 N

Part b)

For the same value of the weight as the surface of moon the acceleration due to gravity of earth must be 1/6 times

so we have

\frac{GM}{r^2} = \frac{GM}{6R^2}

r^2 = 6R^2

r = 2.45 R

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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
sergeinik [125]

Planet Y has rotated by 135.5° through during this time.

To find the answer, we need to know about the relation between angle and radius of orbit.

<h3>What's the expression of angle in terms of radius?</h3>
  • Angle= arc/radius
  • As arc = orbital velocity × time,

            angle= (orbital velocity × time)/radius

  • Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sun
  • So, angle = (√(GM)× time)/radius^3/2
<h3>What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?</h3>
  • Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet X
  • As time = 5 years ( a constant)
  • Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)
  • Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂

   = (4/3)^(3/2) × 88°

   = 135.5°

Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

8 0
2 years ago
A (hypothetical) large slingshot is stretched 4.00 m to launch a 440 g projectile with speed sufficient to escape from Earth (11
RSB [31]

Answer:

(A) 3,449,600 N/m

(B) 62,720 poeple

Explanation:

extension of the sling shot (e) = 4 m

mass of projectile (m) = 440 g = 0.44 kg

speed of projectile (v) = 11.2 km/s = 11,200 m/s

average force a person can exert = 220 N

spring constant (k) = ?

(A) When all the elastic potential energy is converted to kinetic energy

o.5ke^{2} = 0.5 mv^{2}

rearranging the above equation

spring constant (K) = \frac{0.5mv^{2} }{0.5e^{2}}

spring constant (K) = \frac{0.5x0.44x11,200^{2} }{0.5x4^{2}}

spring constant (K) = \frac{27,596,800}{8}

spring constant (K) = 3,449,600 N/m

(B) force required to stretch the slingshot (F) = ke = 3,449,600 x 4 = 13,798,400 N

number of people required = required force / average force per person = 13,798,400 / 220 =62,720 poeple

5 0
3 years ago
A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c
icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

W = F x = \mu N x = \mu\ m\ g\ x

W = 150 \times 6

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

\mu = \frac {F}{m\timesg}

= \frac{150}{40\times 9.8}

= .383

We simply applied the above formulas so that each one part could calculate

3 0
4 years ago
Some instruments differentiate individual quanta of electromagnetic radiation based on their energies. assume such an instrument
lesya [120]
There is a relationship between the energy of a photon and its wavelength. This can be expressed as a mathematical equation shown below:

E = hc/λ
where
h is the Planck's constant equal to 6.62607004 × 10⁻³⁴ m²<span> kg / s
c is the speed of light equal to 3</span>× 10⁸ m/s
λ is the wavelength

3.5×10⁻¹⁶ J = (6.62607004 × 10⁻³⁴ m² kg / s)(3× 10⁸ m/s)/λ
Solving for λ,
λ = 56.8×10⁻⁹ m or<em> 56.8 nm</em>
5 0
4 years ago
The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 241 m the field has
mihalych1998 [28]

Answer:

Explanation:

We shall apply Gauss's Law to find the solution .

flux entering the cube = E X A

= 46.8 X 147.1 X 147.1 Weber

flux going out of cube

= 114 x  147.1 X 147.1  Weber

Net flux going out

= ( 114 x  147.1 X 147.1  - 46.8 X 147.1 X 147.1 )

= 1454101.15 weber .

according to Gauss's law

q / ε₀ = 1.4541 x 10⁶ , q is required charge inside the cube.

q = 8.85 x 10⁻¹² x 1.4541 x 10⁶

= 12.868 x 10⁻⁶ C

= 12.868 μC.

7 0
3 years ago
Read 2 more answers
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