According to all the given requirements
stratosphere x starts at an altitude of 16 km
that means
<span>16≤x
</span> and,as given ,stratosphere ends at an altitude of 30 km:
which means
<span>x≤30</span>
combining both of them we get
<span>16≤x≤30</span>
So the range of the stratosphere is from 16 to 30 km.
Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature
m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).
H = 6.9 kj = 6.9 *1000J = 6900 J
6900 = 200*0.444* (θ₂ - 22)
6900/(200*0.444) = θ₂ - 22
77.70 = θ₂ - 22
θ₂ - 22 = 77.7
θ₂ = 77.7 + 22 = 99.7
So initial temperature before cooling ≈ 100°C . Option C.
Answer:
The spring stretched by x = 13.7 cm
Explanation:
Given data
Mass = 3 kg
k = 120 
Angle
= 34°
From the free body diagram
Force acting on the box = mg sin
⇒ F = 3 × 9.81 × 
⇒ F = 16.45 N ------- (1)
Since box is attached with the spring so a spring force also acts on the box.
= k x
= 120
-------- (2)
The net force acting on the body is given by
Since acceleration of the box is zero so



Put the values from equation (1) & (2) we get
16.45 = 120
x = 0.137 m
x = 13.7 cm
Therefore the spring stretched by x = 13.7 cm
Remember that the total
velocity of the motion is the vector sum of the velocity you would have in
still water and the stream. Always place the vectors carefully to be able to
come up with an accurate sum vector.
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Explanation:
Below is an attachment containing the solution.