We know the equation
weight = mass × gravity
To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.
So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)
Therefore,
100N = mass × 10
mass= 100N/10
mass= 10 kg
Now, all we need are the moon's gravitational field strength and to apply this to the equation
weight = 10kg × (gravity on moon)
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
Answer:
- 5436 J
Explanation:
mass of car, m = 120 kg
radius of loop, r = 12 m
velocity at the bottom (A) = Va = 25 m/s
Velocity at the top(B) = Vb = 8 m/s
Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m
by use of Work energy theorem
Work done by all the forces = change in kinetic energy of the body
Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A
- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)
- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)
- 28224 + Work done by friction = - 33660
Work done by friction = -33660 + 28224 = - 5436 J
B. 0.6 for show
hopefully this works lemme know
So here is my answer. Given that the electric field in <span>a 3.0mm×3.0mm square aluminum wire is 1.0×10−2 V/m, this is how we find the current in the wire.
</span><span>First, we take the distance as 3mm or 0.003m. Using the formula E=V/d, where d is distance, v voltage and E electric field strength, we make V the subject, being V=Ed or 2.2*10^-2*0.003=6.6*10^-5V
</span>Hope this answers your question.
