option b )2004........
How much power is used if a force of 35 newtons is used to push a box a distance of 10 meters in 5 seconds?w=350j
Answer:
How much power is used if 350J of work is done when pushing a box for 5 seconds.
Explanation:
the answer is 70watts
What 2 factors affect the impulse on an object in a collision?
In its American colonies, Spain helped the Catholic Church meet its goal of
finding gold.
gaining territory.
converting people.
achieving glory.
This is from ed please help thank you
Answer: Converting people
Explanation: In Spanish colonies such as Florida, Spanish moved up into the other states to spread their christianity and were successful with doing so.
As we know, the moon is a satellite of our earth, what is the
theoretical period of the moon? The average radius of the
moon's orbit is 3.84 108 m and the mass of the earth is 5.97 x
1024 kg (in hours, G = 6.67 x 10-9 N (m/kg) 3).
Answer:c
Explanation:c
This question involves the concepts of the time period, orbital radius, and gravitational constant.
The theoretical period of the moon is "658 hr".
The theoretical time period of the moon around the earth can be found using the following formula:
[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}[/tex]
where,
T = Time Period of Moon = ?
R = Orbital Radius = 3.84 x 10⁸ m
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
Therefore,
[tex]\frac{T^2}{(3.84\ x\ 10^8\ m)^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\T^2=(9.91\ x\ 10^{-14}\ s^2/m^3)(56.62\ x\ 10^{24}\ m^3)\\\\T=\sqrt{561.34\ x\ 10^{10}\ s^2}[/tex]
T = 2.37 x 10⁶ s[tex](\frac{1\ h}{3600\ s})[/tex]
T = 658 hr
Learn more about the orbital time period here:
https://brainly.com/question/14494804?referrer=searchResults
The attached picture shows the derivation of the formula for orbital speed.
Which type of energies make up the mechanical energy of a roller coaster moving along a track?
Answer:
gravitational potential energy and kinetic energy
Explanation:
A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10s. If the airplane is putting out an average force of 5.8810x10^4 N during this time, what is the average friction force exerted on the airplane by the air?
Given :
A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10 s.
If the airplane is putting out an average force of [tex]5.8810\times 10^4 \ N[/tex].
To Find :
The average friction force exerted on the airplane by the air.
Solution :
Acceleration is given by :
[tex]a = \dfrac{162-120}{2.10}\ m/s^2\\\\a = 20 \ m/s^2[/tex]
Now, force equation is given by :
[tex]F - F_{friction} = ma\\\\F_{friction} = F-ma\\\\F_{friction} = 58810 - (2450\times 20 )\\\\F_{friction} = 9810\ N[/tex]
Therefore, frictional force exerted in the airplane by the air is 9810 N.