The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.
Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
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Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
![R = \frac{6}{2.5} - 2.4 ohm](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7B6%7D%7B2.5%7D%20-%202.4%20ohm)
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
![\frac{3R}{R+3} \times 2.5 = 6](https://tex.z-dn.net/?f=%5Cfrac%7B3R%7D%7BR%2B3%7D%20%5Ctimes%202.5%20%3D%206)
R= 12 ohm
2) circuit Q
![I\times (R+0.1) =V](https://tex.z-dn.net/?f=I%5Ctimes%20%28R%2B0.1%29%20%3DV)
![R+0.1 =\frac{6}{2.5}](https://tex.z-dn.net/?f=R%2B0.1%20%3D%5Cfrac%7B6%7D%7B2.5%7D)
R = 2.3 ohm
Answer:
Power is 1061.67W
Explanation:
Power=force×distance/time
Power=65×9.8×15/9 assuming gravity=9.8m/s²
Power=3185/3=1061.67W
Answer:
Option A is correct.
(The faster object encounters more resistance)
Explanation:
Option A is correct. (The faster object encounters more resistance)
Air resistance depends on various factors:
- Speed of the object
- Cross-sectional area of the object
- Shape of the object
Formula:
![F=\frac{1}{2}C_d\rho A v^{2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B1%7D%7B2%7DC_d%5Crho%20A%20v%5E%7B2%7D)
As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.
south = -(north)
Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)
Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)
Displacement = (4 - 2 + 5 - 5) km north
<u>Displacement = 2 km north </u>