Answer:
[1, 6, -2]
Explanation:
Given the following :
Initial Position of spaceship : [3 2 4] km
Velocity of spaceship : [-1 2 - 3] km/hr
Location of ship after two hours have passed :
Distance moved by spaceship :
Velocity × time
[-1 2 -3] × 2 = [-2 4 -6]
Location of ship after two hours :
Initial position + distance moved
[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]
= [3-2, 2+4, 4-6] = [1, 6, -2]
Answer:
f = 692 N
Explanation:
given data:
f =800N
a =1.2 m s^{2}
m= 90 kg
from newton's second law
net force 
therefore we have from above equation
ma =F - f
putting all value to get force of friction
1.2*90 = 800 - f
f = 692 N
Answer:
h'=0.25m/s
Explanation:
In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).
So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of
. As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.
If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

When solving for r, we get:

so we can substitute this into our volume of a cone formula:

which simplifies to:


So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

Which simplifies to:

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)
So we get:

Now we can substitute the provided values into our equation. So we get:

so:

Answer:
New pressure is 0.534 atm
Explanation:
Given:
Initial volume of the gas, V₁ = 250 mL
Initial pressure of the gas, P₁ = 1.00 atm
Initial temperature of the gas, T₁ = 20° C = 293 K
Final volume of the gas, V₂ = 500 mL
Final pressure of the gas = P₂
Final temperature of the gas, T₁ = 40° C = 313 K
now,
we know for a gas
PV = nRT
where,
n is the moles
R is the ideal gas constant
also, for a constant gas
we have
(P₁V₁/T₁) = (P₂V₂/T₂)
on substituting the values in the above equation, we get
(1.00 × 250)/293 = (P₂ × 500)/313
or
P₂ = 0.534 atm
Hence, the <u>new pressure is 0.534 atm</u>