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!
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.
I think it’s false bc A volcano is not in the interior of the earth
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/
, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut +
a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s =
a
=
x 14 x 
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
<span>4.5 m/s
This is an exercise in centripetal force. The formula is
F = mv^2/r
where
m = mass
v = velocity
r = radius
Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop.
Let's determine the force we get from gravity.
0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N
Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N
Plug known values into formula.
F = mv^2/r
13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m
6.88064 kg m^2/s^2 = 0.34 kg V^2
20.23717647 m^2/s^2 = V^2
4.498574938 m/s = V
Rounding to 2 significant figures gives 4.5 m/s
The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>