Answer:
mantle convection is the very slow creeping motion of earths solid silicate mantle caused by convection currents carrying heat from the interior to the planet's surface.
Answer:
Hello your question has some missing parts attached below is the complete question
answer : 4 μm
Explanation:
since the scale bar works the same way as a scale on a map , each bar will therefore represent 1 μm and the mature parent cell's is about 4 times the labeled value hence the Mature parent cell diameter will approximately be : 4 μm
The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.
The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.
So;
Period(T) = 54 minutes or 3240 seconds
Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz
Learn more: brainly.com/question/14588679
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so

=> 

integrating both sides

Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have

![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 
