Mass is the same, weight is less
<h3>What is the Weight and mass on Moon ?</h3>
As we know that the mass of the object is the measurement of the quantity of the matter that is present in it
So here we can say that if the mass of the object is m then its total quantity of the matter that is present in it is given as
mass = (density) × (volume)
Now for the weight of the object is defined as the force of gravity due to planet
Fg = mg
so the weight of the object is depending on the acceleration due to gravity of the planet
As we know that the gravity of moon is smaller than the gravity of the earth so here weight on the moon will be smaller than the weight on the Earth
Learn more about Weight on Moon here:
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Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
Answer:
= 0.7 A, = 1.3 A and ε = 7.4 V
Explanation:
From the given circuit, applying Kirchhoff's rule;
Ammeter reading, = 2 A
⇒ = + = 2 A
Dividing the circuit to loops 1 and 2.
a. From loop 1,
15 + 7 - 5 = 0
15 + 7 - 10 = 0 (since = 2 A)
7 - 5 = 0
= 0.7 A
But, = +
⇒ 2 = 0.7 +
= 1.3 A
b. From loop 2,
ε + 2 - 5 = 0
ε + 2 - 10 = 0
ε + 2.6 - 10 = 0
ε - 7.4 = 0
ε = 7.4 V
Therefore, = 0.7 A, = 1.3 A and ε = 7.4 V.