Answer:
T1 = 417.48N
T2 = 361.54N
T3 = 208.74N
Explanation:
Using the sin rule to fine the tension in the strings;
Given
amass = 42.6kg
Weight = 42.6 * 9.8 = 417.48N
The third angle will be 180-(60+30)= 90 degrees
Using the sine rule
W/Sin 90 = T3/sin 30 = T2/sin 60
Get T3;
W/Sin 90 = T3/sin 30
417.48/1 = T3/sin30
T3 = 417.48sin30
T3 = 417.48(0.5)
T3 = 208.74N
Also;
W/sin90 = T2/sin 60
417.48/1 = T2/sin60
T2 = 417.48sin60
T2 = 417.48(0.8660)
T2 = 361.54N
The Tension T1 = Weight of the object = 417.48N
V=IR can be changed to V/R=I so 10V/2 ohms = 5amps so 5 amps is your answer boss
Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm
