Well this all depends on the region you would like to know about. One biome would be The Tundra. This biome is a very bitter cold. Some times the temperature can drop to -45f! So your answer more than likely would be Tundra.
Have a wonderful day user!
Answer:
1110 N
Explanation:
First, find the acceleration.
Given:
Δx = 300 m
v₀ = 85.5 km/h = 23.75 m/s
v = 0 m/s
Find: a
v² = v₀² + 2aΔx
(0 m/s)² = (23.75 m/s)² + 2a (300 m)
a = -0.94 m/s²
Find the force:
F = ma
F = (1180 kg) (-0.94 m/s²)
F = -1110 N
The magnitude of the force is 1110 N.
Answer:
h=17357.9m
Explanation:
The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.
To calculate this, you need to use the barometric formula:

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.
Furthermore, the specific gas constant is defined by:

Therefore yo can write the barometric formula as:

at the surface of the planet (h =0) the pressure is ![P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}](https://tex.z-dn.net/?f=P_0%5B%5Ctex%5D.%20The%20pressure%20at%20the%20height%20requested%20is%20half%20of%20that%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%3D%5Cfrac%7BP_0%7D%7B2%7D)
applying to the previuos equation:

solving for h:
h=17357.9m
Answer:
(a) The least time required for the rotation is 50 seconds
(b) The corresponding value of α₁ is 1.6 rad/s²
Explanation: Please see the attachments below
Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately