PLS ANSWER ASAP

A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

AVprozaik [17]

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

6 0

3 years ago

a baby carriage is sitting at the top of a hill that is 21 m high. the carriage with the baby has a mass of 1.5 kg. the carriage

Serga [27]

Height of baby carriage from ground = 21m

Mass of carriage with baby = 1.5 kg

The carriage has potential energy by virtue of its height.

Potential energy = mgh = 1.5×10×21 = 315 J

Hence, potential energy of the carriage is 315 Joule.

7 0

3 years ago

"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 second

Leni [432]

<span>Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculations of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:

d = v0t + at^2/2

We cancel the term with v0 since it is initially at rest,

d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2

</span>

6 0

2 years ago

Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3

Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

$k_{1}=20\ N/m$

$k_{2}=30\ N/m$

$k_{3}=15\ N/m$

$k_{4}=20\ N/m$

$k_{5}=35\ N/m$

According to figure,

$k_{2}$ and $k_{3}$ is in series

We need to calculate the equivalent

Using formula for series

$\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}$

$k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}$

Put the value into the formula

$k=\dfrac{30\times15}{30+15}$

$k=10\ N/m$

k and $k_{4}$ is in parallel

We need to calculate the k'

Using formula for parallel

$k'=k+k_{4}$

Put the value into the formula

$k'=10+20$

$k'=30\ N/m$

$k_{1}$ ,k' and $k_{5}$ is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

$k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}$

Put the value into the formula

$k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}$

$k_{eq}=8.93\ N/m$

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0

3 years ago

When running your engine, you cause debris, rocks and propeller blast to be directed towards people or other aircraft. Is this c

beks73 [17]

Answer:

Yes, it is reckless. This is because it is the responsibility of the pilot to make sure that the direction of the propeller blast is away from people or other aircraft and in a safe direction.

Explanation:

Yes, it is reckless to let the propeller blast face people and other aircraft. This is because it is the responsibility of the pilot to make sure that the direction of the propeller blast is away from people or other aircraft and in a safe direction. People and other aircraft can be injured by the debris and the rocks that are scattered by the engine of the aircraft.

3 0

2 years ago