The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
Northern Lights ( Aurora Borealis)
Explanation:
When the electricaly charged sunspot gases (they are named a solar wind) escape the sun's chromosphere and penetrates from the earth magnetic sheild which is called earth's magnetosphere then upon there interaction with atoms and molecules of our atmosphere there are little bursts of photons in the form of light which made up these northern lights.
Answer:
The weight of body is 1.3040 gram.
Explanation:
Given that,
The weight y of a fiddler crab is directly proportional to the 1.25 power of the weight x of its claws.
Suppose a crab with a body weight of 1.8 gram has claws weighing 1.1 gram.
Estimate the weight of a fiddler crab with claws weighing 0.85 gram.
Determine the weight of crab body
We need to calculate the value of proportional constant
Put the value into the formula
We need to calculate the crab weight
Here, x = 0.85 g
Put the value into the formula
Hence, The weight of body is 1.3040 gram.
Answer:
4.399 Nm
Explanation:
The maximum Torque on a coil is given as,
τ = BNIA...................... Equation 1
Where τ = Maximum torque exerted on the coil, B = Magnetic Field, N = Number of turns, I = Current, A = Area.
Given: N = 45.5 Turns, B = 0.49 T, I = 26.7 mA = 0.0267 A,
A = πr², Where r = radius of the coil, r= 4.85 cm = 0.0485 m
A = 3.14(0.0485)²
A = 7.39×10⁻³ m².
Substitute into equation 1
τ = 45.5×0.49×26.7×7.39×10⁻³
τ = 4.399 Nm
Q7. A fizzing sound was overheard shadowed by the rushing out
of bubbles from the bottle dipped in hot water. There was also a sound perceived
in the bottle to be found in cold water but not as much as in bottle A.
Q8. There was amassed gas inside the bottle.
Q9. The hot water surges the temperature of the soda drink
inside the bottle. As the temperature increases, more gas is accumulated inside
the bottle. This reasons the fizzing sound.
Q10. The observation in the bottle of cooking oil is not the
same as in the soda drinks.
<span>Q11. There was little gas out in the bottle of cooking oil for
the reason of its structure. We know that soda drink is carbonated. The great
temperature free the gas from the soda drinks.</span>
