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just olya [345]

3 years ago

13

The earth takes 1 year to revolve around the sun at 1 A.U. distance (an astronomical unit = 93,000,000 miles). If a planet were

4 A.U. from the sun, how many years would it take to make 1 orbit?

1 answer:

inessss [21]3 years ago

4 0

Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis of its orbit. In equation form, P^2 / a^3 = constant. For the earth, (1 y)^2 / (1 AU)^3 = 1, so the constant is 1. For a planet 4 AU from the sun:
P^2 / (4 AU)^3 = 1P^2 = 64P = 8 years

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Differentiate between the factor(s) that affect inertia and momentum.

Oksanka [162]

Answer:

Inertia relates directly to an object's <em>mass</em>. It is its inherent property and expresses how much force is needed to bring the object to a certain level of acceleration. It is the sole factor.

Momentum relates proportionally to two factors: <em>mass</em> and <em>velocity</em>. As such, momentum is not an inherent property of an object but a measure dependent on it current velocity, in addition to its mass.

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3 years ago

Magnetism is a ?<br> A) FORCE<br> B) POWER<br> C) FORM OF ENERGY<br> D) FORM OF ELECTRICITY

Ksivusya [100]

I believe this answer is Force

3 0

2 years ago

Read 2 more answers

3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.

Novay_Z [31]

Answer:

A. The bullet with 0.006kg has more energy

B. When the mass is doubled the kinetic energy increases

Explanation:

Kinetic energy increases when mass increases

kinetic energy increases when velocity increases

6 0

3 years ago

The empire state building is 1,450 feet tall. King Kong weighs 20 tons and he climbs to the very top. If he jumps off the top, w

lozanna [386]

Answer:velocity= $93.12m/s$

Explanation:

We shall use conservation of energy to solve this problem

we have

Potential energy of King Kong at top of building = His kinetic energy at the bottom

Potential energy of an object = $m\times g\times h$ = 20000 kg

Where m is mass of an object(20000 kg)

g is accleration due to gravity = 9.81 $m/s^{2}$

h is the height above the surface of earth = 1450 feet = 441.96 meters

Applying values we get

$(P.E)_{TOP}=(20000\times 9.81\times 441.96) Joules\\\\(P.E)_{TOP}= 86712.552KiloJoules$ ......................(i)

Now Kinetic energy is given by

$(K.E)_{Bottom}=\frac{1}{2}mv^{2}$ .......................(ii)

Equating i and ii we get

86712.552 = $\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2E}{m}}$

Applying values we get

$v=\sqrt{\frac{2\times 86712.552\times 10^{3}}{20\times 10^{3}}}\\\\v= 93.12m/s$

3 0

3 years ago

A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago