A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
Answer:
hence option A is correct
Explanation:
heat required from -9°C to 0°C ice = mass × specific heat of ice ×change in temperature
heat required from -9°C to 0°C ice = 7×2100×9 =132300 J =0.1323 MJ
( HERE SPECIFIC HEAT OF ICE IS A CONSTANT VALUE OF 2100
J/(kg °C )
heat required from 0°C ice to 0°C water = mass× specific heat of fusion of ice
= 7×3.36×10^5
= 2.352 × 10^6 J
= 2.352 MJ
TOTAL HEAT ENERGY REQUIRED = 0.1323 MJ +2.352 MJ
= 2.4843 MJ
hence option A is correct
Check the attached file for the answer.
Answer:
t=40s,
Explanation:
If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river, if the water flows downstream at a rate of 1.5m/s, is most nearly:
from the question the swimmer will have a velocity which is equal to the sum of the speed of the water and the velocity to swi across the bank
Vt=v1+v2
the time is takes to swim across the bank will be
DY=Dv*t
DY=distance across the bank
Dv=ther velocity of the swimmer across the bank
t=20/ 0.5m/s,
t=40s, time it takes to swim across the bank
velocity is the rate of displacement
displacement is distance covered in a specific direction
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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