<u>Answer:</u>
<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.
<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.
<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.
Answer: 288.8 m
Explanation:
We have the following data:
is the time it takes to the child to reach the bottom of the slope
is the initial velocity (the child started from rest)
is the angle of the slope
is the length of the slope
Now, the Force exerted on the sled along the ramp is:
(1)
Where
is the mass of the sled and
its acceleration
In addition, if we draw a free body diagram of this sled, the force along the ramp will be:
(2)
Where
is the acceleration due gravity
Then:
(3)
Finding
:
(4)
(5)
(6)
Now, we will use the following kinematic equations to find
:
(7)
(8)
Where
is the final velocity
Finding
from (7):
(9)
(10)
Substituting (10) in (8):
(11)
Finding
:

<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:
F = G(Mm / (4(pi)*r^2))
The above expression gives the force that you feel on the earth's surface, as it is today!
Let us now double the mass of the earth and decrease its diameter to half its original size.
This is the same as replacing M with 2M and r with r/2.
Now the gravitational force (F' ) on the new earth's surface is given by:
F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F
So:
F' = 8F
This implies that the force that you would feel pulling you down (your weight) would increase by 800%!
You would be 8 times heavier on this "new" earth!</span>
Answer:
v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²
Explanation:
This is a missile throwing exercise.
On the x axis there is no acceleration so the velocity on the x axis is constant
v₀ₓ = 10 m / s
On the y-axis velocity is affected by the acceleration of gravity, let's use the equation
v_y =
- g t
at the highest point of the trajectory the vertical speed must be zero
v_y = 0
therefore the velocity of the body is
v = (10 i ^ + 0j ^) m / s
the acceleration is
a = (0 i ^ - g j⁾
a = (0i ^ - 9.8 j ^) m / s²