Answer:
h ’= 12,768 cm
Explanation:
For this exercise let's use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
the magnification equation is
m = h '/ h = -q / p
let's find the distance to the object
1 / p = 1 / f- 1 / q
1 / p = 1/20 - 1 / (- 37.5)
1 / p = 0.076666
p = 13.04 cm
now let's use the magnification equation
h ’= - q / p h
let's calculate
h ’= - (-37.5) / 13.04 4.44
h ’= 12,768 cm
<span>We know that pressure is the force applied into a surface, in our case the wall of the room, so then first we will calculate the surface of this wall:
S = 2.2 * 3.2 = 7.04 m2
Then we also know the atmospheric pressure in normal conditions is 1 atm. That is the same 1 atm = 101325 Pascals or 101325 N/m2
Now we need to use the formula : P = F/S where P is pressure, F is force and S is surface to calculate the force:
F = P * S = 101325 * 7.04 = 713,328 Newtons
Conclusion: the force acts on the wall due the air inside the room is 713,328 N</span>
According to my guesses, he should have swung the pendulam bob and noted its time period. In order to observe the effect of mass, he would have repeated the experiment with varied pendulam bobs. Hope this helped!
Answer:
Explanation:
Acceleration on a VT graph is the slope of the line at the given point. We can find the slope at 3 with Δy/Δx. This gives us (4-2)/(3-(-3)) which works out to be -3m/s^2
Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.