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pearlsawme

pearlsawme

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pearlsawme
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  • a. Charge Q3 has the largest magnitude of all. Considering the equipotenttial of 5V, the distance of this curve is greater for Q3. Hence its charge must be high in magnitude. b. The force on an electron at g points to the top of the page. Fal…
  • a. Charge Q3 has the largest magnitude of all. Considering the equipotenttial of 5V, the distance of this curve is greater for Q3. Hence its charge must be high in magnitude. b. The force on an electron at g points to the top of the page. Fal…
  • angular momentum of the mass is 4.61 kg m² /s You apply a force and it is the tension in the string which provides the centripetal force Tension = mv²/r = 0.8*3.6²/1.6 =6.48 N. ( why do you use "mg " here ?) New tension in th…
  • 0.0047 kg/m^3*s^1 False. If it is written as 4.7E-6 g/ (m^3*s^1) then it is true. 0.0047 kg*m^-3*s^-1 True. 4.7 g*cm^-3*s^-1 ( magnitude is wrong ) False 4.7E-6 g/ cm^3*s^1 False. If it is written as 4.7E-6 g/ (cm^3*s^1) then…
  • 1 Points x and z are at the same temperature The product Px.Vx = PzVz 2. In process 1 , volume increases from x to y In process 2 volume decreases from z to x . 3 The work done on the gas during process I. is negative since volume inc…
  • 1 Points x and z are at the same temperature The product Px.Vx = PzVz 2. In process 1 , volume increases from x to y In process 2 volume decreases from z to x . 3 The work done on the gas during process I. is negative since volume inc…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • One has to use a table or calculator to find the angles . But you can memorize certain trigonometric values of certain familiar angles like 30 ,45 60 and 90. which will be of much use full .
  • One has to use a table or calculator to find the angles . But you can memorize certain trigonometric values of certain familiar angles like 30 ,45 60 and 90. which will be of much use full .
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • In the direction of x the net force is 10.1 cos 25˚ + 14.2 cos 170˚ + 11.3 cos 248˚ = -9.04 N In the direction of y , the net force is 10.1 sin 25˚ + 14.2 sin 170˚ + 11.3 sin 248˚ = - 3.743 N Resultant of these two forces is √ { (-9.0…
  • 1 Points x and z are at the same temperature The product Px.Vx = PzVz 2. In process 1 , volume increases from x to y In process 2 volume decreases from z to x . 3 The work done on the gas during process I. is negative since volume inc…
  • 1 Points x and z are at the same temperature The product Px.Vx = PzVz 2. In process 1 , volume increases from x to y In process 2 volume decreases from z to x . 3 The work done on the gas during process I. is negative since volume inc…
  • 0.0047 kg/m^3*s^1 False. If it is written as 4.7E-6 g/ (m^3*s^1) then it is true. 0.0047 kg*m^-3*s^-1 True. 4.7 g*cm^-3*s^-1 ( magnitude is wrong ) False 4.7E-6 g/ cm^3*s^1 False. If it is written as 4.7E-6 g/ (cm^3*s^1) then…
  • angular momentum of the mass is 4.61 kg m² /s You apply a force and it is the tension in the string which provides the centripetal force Tension = mv²/r = 0.8*3.6²/1.6 =6.48 N. ( why do you use "mg " here ?) New tension in th…