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2 years ago

A round asteroid has a surface gravity of .0288 m/s^2 if the mass of the asteroid is 1.10 x10^18 kg what is its radius

Physics

1 answer:

Oduvanchick [21]2 years ago

6 0

Answer:

r = 50.47 x 10³ m = 50.47 km

Explanation:

Using the formula for the acceleration due to gravity:

$g = \frac{Gm}{r^2}$

where,

g = acceleration due to gravity on the surface of asteroid = 0.0288 m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m= mass of asteroid = 1.1 x 10¹⁸ kg

r = radius of asteroid = ?

Therefore,

$0.0288\ m/s^2 = \frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{r^2} \\\\r = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{0.0288\ m/s^2}}$

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

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2 years ago

Conversion of one form of energy into another form is

Artist 52 [7]

Answer: energy transformation

5 0

2 years ago

Read 2 more answers

A 1-kilogram parcel of air is at 35°c and contains 7 grams of water vapor. What is the relative humidity?

Nitella [24]

Answer:

20%

Explanation:

Relative Humidity (%) = (water vapor content÷water vapor capacity) × 100

=(7÷35)×100

=(0.2)×100

=20%

According to the Temperature-Water Vapor Capacity Table, the water capacity at 35 °C is 35 grams.

Water Vapor Capacity: The amount of water (grams) which air can hold at a given temperature.

Water Vapor Content: The amount of water vapor actually present in the air.

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3 years ago

Find the distance covered by a car in 10 second, if it is moving at the speed of 20m/s

labwork [276]

Answer:

solution

we know that .

s=ut+½at²

now, putting the values in the second equation of motion ,

we get,

S=20×10+1/2×(10)²

S=200+1×100

S=200+100

S=300m

the distance covered by given body or object in 10 seconds is 300 m.

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3 years ago

Problem 13.175 A 1-kg block B is moving with a velocity v0 of magnitude as it hits the 0.5-kg sphere A, which is at rest and han

Svetradugi [14.3K]

Explanation:

Mass of the Block, Mb = 1.0 kg Initial Velocity of Block, Vb = 2.0 m/s

Mass of the Ball, Ma = 0.5 kg Coefficient of Kinetic Friction, μ = 0.6

Coefficient of Restitution, e = 0.8 Gravity = 9.81 m/s^2

Sum of the forces of the X-axis components and Y-axis components are:

∑ Fx = Ff = Mb × a

Equation for frictional force,

Ff = μ × N

∑ Fy = N - mb × g = 0

Note:

N = mb × g

Therefore, to solve for the acceleration, we have:

μ × Mb × g = Mb × a

a = μ × g

= 0.6 × 9.81

= 5.88 m/s^2

Therefore, ratio of velocities using the coefficient of restitution, e:

e = (Vb2 – Va1)/ (Va – Vb)

Note: Va = zero (initially at rest)

Va2 – Vb2 = 0.8 × ( 0 - 2 m/s)

Vb2 – Va2 = -1.6 m/s

Vb2 = a2 – 1.6

Using Conservation of Momentum Equation:

Ma × Va + Mb × Vb = Ma × Va2 + Mb × Vb2

0 + 1kg × 2 m/s = 0.5 kg × Va2 + 1 kg × Vb2

Vb2 = 2 - 0.5 × Va2

Substitute in Vb2,

1.5 × Va2 = 3.6

Va2 = 2.4 m/s

Vbs = 0.8 m/s

mgh = 0.5 × Mv2

h = v2/ 2g

h = 0.294 m

Using equation of motion,

Vf^2 = Vo^2 + 2a × S

Given:

Vf = 0 m/s

a = 5.88 m/s^2

0 = 0.82 + 2(5.88) × ΔS

Δx = 0.0544 m

4 0

3 years ago