Answer:
B. Geosphere
A. Biosphere
A. Atmosphere
Explanation:
Volcanic eruptions occurs within the Geosphere. The geosphere is the rock solid earth make up of rocks that extends into the deep interior.
Magma formed deep within the crust rises to elevated parts and finally erupts as lava on the surface. When they cool, they solidify to form volcanic rocks.
The volcanic eruptions affects the biosphere significantly. The biosphere is the portion of the earth where all life forms exists.
Gases and ash spewed during an eruption into the atmosphere causes severe changes to weather and leads to pollution. The atmosphere is the gaseous envelope round the earth.
The distance of the galaxy is 32.86 Mpc.
Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.
Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have
D = v/H₀
Substituting the values of the variables into the equation, we have
D = 2300 km/s ÷ 70 km/s/Mpc
D = 32.86 Mpc
So, the distance of the galaxy is 32.86 Mpc
Learn more about hubble law here:
brainly.com/question/18484687
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>
the velocity is Zero when the projectile reach in the maximum altitude:
When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>
R=Range
**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile
2B=60°
B=30°
