Yes, that's right. It's the 'Planck' length, not the 'Planet' length.
You could easily find these with a web search. But in gratitude
for the bountiful 5 points, I've saved you the trouble.
AND guess what ! By doing that, I learned something, and
you didn't.
Speed of light (c): 299,792,458 meters per second
Gravitational constant (G): 6.67 x 10⁻¹¹ newton-meter²/kilogram²
Planck's Konstant (h): 6.63 x 10⁻³⁴ joule-second
Planck Length: 1.6 x 10⁻³⁵ meter
(about 10⁻²⁰ the size of a proton)
Planck Time: 10⁻⁴³ second
(about the time it takes to travel
a Planck Length at the speed of light)
Answer:
Becomes softer as the temperature rises!
Explanation:
Newton's first law states that an object at rest will stay at rest && an object in motion will remain in motion unless acted on by another force. With that being said you feel tossed around because you are initially at rest but when you ride the roller coaster you take it's course of motion.
Hoped this helped!
Answer:
v = 719.2 m / s and a = 83.33 m / s²
Explanation:
This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is
v - v₀ = 
ln (M₀ / M)
where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket
In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s
m_fuel = 75 10
m_fuel = 750 kg
As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is
M = 3000 -750 = 2250 kg
let's calculate
v - 0 = 2500 ln (3000/2250)
v = 719.2 m / s
To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is
Push = v_{e} dM / dt
let's calculate
Push = 2500 75
Push = 187500 N
If we use Newton's second law
F = m a
a = F / m
let's calculate
a = 187500/2250
a = 83.33 m / s²
Answer:
a) —0.5 j m/s
b) 4.5 i + 2.25 j m
Explanation:
<u>Givens:</u>
v_0 =3.00 i m/s
a= (-3 i — 1.400 j ) m/s^2
The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :
<em>v_x = v_0 + at = 0 </em>
(3.00 i m/s) + (-3 i m/s^2)t=0
t = (3 m/s)/-3 i m/s^2
t = -1 s
Therefore the particle reaches the maximum x-coordinate at time t = 1 s.
Part a The velocity-of course- is all in the y-direction,therefore:
v_y =v_0+ at
We have that v_0 = 0 in the y-direction.
v_y = (-0.5 j m/s^2)(1 s)
= —0.5 j m/s
Part b: While the position of the particle at t = 1 s is given by:
r=r_0+v_0*t+1/2*a*t^2
Where r_0 = 0 since the particle started from the origin.
Its position at t = 1 s is then given by :
r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2
=4.5 i + 2.25 j m
