Answer:
a) the Tunguska meteoric impact
Explanation:
The Tunguska Event, sometimes known as the Tungus Meteorite is thought to have resulted from an asteroid or comet entering the earth's atmosphere and exploding. The event released as much energy as fifteen one-megaton atomic bombs. As well as blasting an enormous amount of dust into the atmosphere, felling 60 million trees over an area of more than 2000 square kilometres. Shaidurov suggests that this explosion would have caused "considerable stirring of the high layers of atmosphere and change its structure." Such meteoric disruption was the trigger for the subsequent rise in global temperatures
According to Vladimir Shaidurov of the Russian Academy of Sciences, the apparent rise in average global temperature recorded by scientists over the last hundred years or so could be due to atmospheric changes that are not connected to human emissions of carbon dioxide from the burning of natural gas and oil.
A a wire, a wire is connected between the positive and negative ends of the eletrical energy source and connects all the components in between.
Answer:
it would take 3.26 seg for the stone to fall to the water
Explanation:
If we ignore air friction then:
h=h₀ + v₀*t -1/2*g*t²
where
h= coordinates of the stone in the y axis ( height of the stone relative to the surface of the water )
h₀ = initial coordinates of the stone ( height of the cliff relative to the surface of the water = 52 m )
v₀ = initial <u>vertical </u>velocity = 0 ( since the ball is kicked horizontally , has only initial horizontal velocity , and has 0 vertical velocity )
t = time to reach a height h
g = gravity = 9.8 m/s²
since v₀ =0
h= h₀ - 1/2*g*t²
h₀ - h = 1/2*g*t²
t= √[2(h₀ - h)/g]
when the stone hits the ground h=0 ( height=0) , then replacing values
t=√[2(h₀ - h)/g]=√[2(52 m- 0 m )/(9.8m/s²)] = 3.26 seg
t= 3.26 seg
it would take 3.26 seg for the stone to fall to the water
The magnitude of static friction force is f_s = 842.8 N
Explanation:
Write down the values given in the question
The wheel of a car has radius r = 0.350 m
The car applies the torque is τ = 295 N m
It is said that the wheels does not slip against the road surface,
Here we apply a force of static friction,
It can be calculated as
Frictional force f_s = τ / r
= 295 Nm / 0.350 m
f_s = 842.8 N
So Hooke's law says that that law is proportional to how much I stretch the spring. Alright. So f=kx<span>. x is the length of the spring now minus its length when it's relaxed and nobody's pulling on it. k is a constant called the spring constant.</span>
