Longitudinal waves travel through a series of ___ and ___ PlEASE HELP FAST!!!!!!
1 answer:
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The rocket travelled a maximum height at 1.0102 km.
Given,
The acceleration of a rocket (a) = 12 m/s²
The altitude of the rocket (s) = 0.50 km = 0.5×10³m
The maximum height of the rocket (h) = ?
Solution,
A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.
The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²
(a) = Δv/Δt
Where , Δv is change in velocity and Δt is change in time.
The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.
(v)= Δx/Δt
Where,Δx is the change in position and Δt is change in time & v is velocity.
Therefore we know the equation of motion is written as,
v² = u² +2as
Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.
Then putting the value ,
v² = 0 + ( 2× 10 × 0.5×10³)m/s
v² = m/s
v = 100 m/s
Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,
v² = u² - 2(g)h
h = (v²- u² ) / 2g
h = 10,000/2×9.8
h = 510.2 m
So that the rocket travelled the maximum height ,
(h)= (0.5 km + 510.2m)
(h) = 1.0102 km
Hence, the rocket travelled at the maximum height h is 1.0102 km
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Answer:
3.14946 rad/s
Explanation:
= Intial moment of inertia
= Final moment of inertia
= Initial angular velocity
= Final angular velocity =
In this system the angular momentum is conserved
The angular velocity when the diver left the board is 3.14946 rad/s
The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x Kg
Linear speed is the product of angular speed and the maximum displacement of the particle. That is,
V = Wr
Where
- V = Linear speed
- W = Angular speed
- r = Radius
Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed
W = 2/T
W = (2 x 3.143) / (365.26 x 24)
W = 6.283 / 876624
W = 7.2 x Rad/hr
The Earth’s average orbital speed V = Wr
V = 7.2 x x 149.6 x
V = 107225.5 kilometers per hours.
b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law
M = (4) / G
M = (4 x 9.8696 x 3.35 x ) / (6.67 x
x 7.68 x
<em>)</em>
<em>M = 1.32 x </em> / 51.226
M = 2.58 x Kg
Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x Kg
Learn more about Orbital Speed here: brainly.com/question/22247460
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