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1 year ago

From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:

Physics

1 answer:

bazaltina [42]1 year ago

7 0

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

$T^2\propto a^3\\\\T^2=ka^3$

It is mentioned that, an asteroid with an orbital period of 8 years. So,

$(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU$

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

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The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at

elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

a) The point at which particle stop, it's velocity = 0 m/s

So dx/dt = 0

0 = 0- 12t = -12t

So when time t= 0, velocity = 0 m/s

So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

$x=4.00 - 6.00*0^2 = 4m$

SO at x = 4m the particle is (momentarily) stop

c) We have $x=4.00 - 6.00t^2$

At origin x = 0

Substituting

$0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}$

t = 0.816 seconds or t = - 0.816 seconds

So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

5 0

2 years ago

Which factors could be potential sources of error in the experiment? check all that apply.

Vadim26 [7]

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

6 0

2 years ago

Read 2 more answers

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the

zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂= L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0

2 years ago

Read 2 more answers

A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki

Anna71 [15]

Based on the direction of propagation compared to direction of vibration, waves are classified into:
1- Transverse waves: The direction of propagation of the wave is perpendicular to the direction of vibration of the medium particles.
2- Longitudinal waves: The direction of propagation of the wave is the same as the direction of vibration of the medium particles.

For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave

6 0

2 years ago

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground = 46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt² ..............1

here d is distance and g is acceleration and t is time

so when object falling it will be

h = 4.9 t² ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)² ...................3

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

so height is 69.68 m

6 0

2 years ago