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Amiraneli [1.4K]
3 years ago
7

SOMEONE HELP PLEASE ASAP I BELIEVE THE ANSWER IS C OR D

Physics
1 answer:
Serhud [2]3 years ago
4 0
You are correct the answer would be C

hope i helps
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What is Virtual Lab and what does people use it for and why they use it?
Dafna1 [17]

The Virtual Laboratory is an interactive environment for creating and conducting simulated experiments: a playground for experimentation. It consists of domain-dependent simulation programs, experimental units called objects that encompass data files, tools that operate on these objects

7 0
3 years ago
You drop a rock from rest from the top of a tall building.1)how far has the rock fallen in 2.60 s?
Bess [88]
Answer:
distance = 33.124 meters

Explanation:
To solve this question, we will use one of the equations of motion which is:
s = ut + 0.5a * t^2
where:
s is the distance that we want to get
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/sec^2
t is the time = 2.6 sec

Substitute with the givens in the equation to get the distance as follows:
s = ut + 0.5a * t^2
s = (0)(2.6) + 0.5(9.8)(2.6)^2
s = 33.124 meters

Hope this helps :)
7 0
3 years ago
How does a savanna differ from a grassland?
Pepsi [2]

Answer:

B. Savannas have shrubs and isolated trees, while grasslands contain grasses.

8 0
3 years ago
Read 2 more answers
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef
inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0
3 years ago
How do clouds become stars ?
kenny6666 [7]
Stars form inside relatively dense concenstrations of interstellar gas and dust known as molecular clouds.





hope it helps
3 0
3 years ago
Read 2 more answers
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