Answer:
we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances
Explanation:
given data
base = 3.60 m
speed u = 8 m/s
height = 1.70 m
to find out
check change in speed
solution
we know here formula for v that is
v² = u² - 2gh ............1 for upward speed
v² = u² + 2gh ............2 for projected speed
so here put all value and find v with h = 3.60 - 1.70 = 1.9 m
v² = 8² - 2(9.8) 1.9 = 26.76
v² = 8² + 2(9.8) 1.9 = 101.24
v = 5.173 m/s ..............3
v = 10.061 m/s ...................4
so change in speed form 3 and 4 equation
change in speed = v - u = 8 - 5.173 = 2.827 m/s .................5
change in speed = v - u = 10.061 - 8 = 2.061 m/s ..................6
so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
I would believe that this is false.
Answer:
3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.
Explanation:
The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =
A = 2πr(r + h)
Where h = height of the can = 12cm
r = radius of the can = 6.5cm/2 = 3.25cm
r = diameter /2
A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²
Atmospheric pressure, P = 101325Pa = 101325 N/m²
F = P × A
F = 101325 ×0.031.
F = 3141N. Or 3.1 ×10³ N.
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
<u>v = 1.7 m/s</u>
