Answer:
-1
Explanation:
Electrons have a negative charge and protons have a positive charge. (+11) + (-12) = -1
Answer:28.8 knots
Explanation:
The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.
First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.
The distance between them is described with Pythogorean theorem, which is
and when we replace the values A and D, we find Distance (D) to be 182 miles.
Now, let's make the notations clear. The velocity of A and B is notated as
and
. The rate of change of distance is also notated as
. Now, we have to find
from the Pythogorean theorem. If we derive the Pythogorean expression
, we would have:

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:
and the answer is 28.6 knots.
Answer:
the questions they could ask would be:
how much will the temperature change?
how much pressure will there be?
how durable does the equipment need to be?
how do we know if we will run into a water pocket or a magma pocket?
Explanation:
Answer:
W / A = 39200 kg / m²
Explanation:
For this problem let's use the equilibrium equation of / newton
F = W
Where F is the force of the door and W the weight of water
W = mg
We use the concept of density
ρ = m / V
m = ρ V
The volume of the water column is
V = A h
We replace
W = ρ A h g
On the other side the cylinder cover has a pressure
P = F / A
F = P A
We match the two equations
P A = ρ A h g
P = ρ g h
P = 39200 Pa
The weight of the water column is
W = 1000 9.8 4 A
W / A = 39200 kg / m²
Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)

![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)