Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Hotter ocean tempatures mean more moisture in the dense air mass
The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.
Answer:
The current in the circuit increases
Explanation:
The ohm's law states that the potential across a circuit is proportional to the current in the circuit.
V ∝ I
Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.
The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes
V = IR
According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.
So, when there is an increase in the voltage, the current on the circuit increases.
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
(c)
Now we will consider the downward motion and use the third equation of motion:
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
