Answer:Wrap more turns of wire around the nail.
Explanation:The power of an electromagnet is directly proportional to the amount of electricity passing through that electromagnet.
Now,if they increase the number of rounds of the copper coil,the electricity conducting surface amount will also increase.
And that's why the electricity passing through that electromagnet will increase along with the electromagnetic power of that electromagnet.
• Increasing the battery voltage :
If they increase the battery voltage,the electricity which is passing through that electromagnet will also increase in amount .
And as discussed previously,the increasing amount of electricity will also increase the electromagnetic abilities of the electromagnet.
Answer:
time is 0.42 sec
Explanation:
Given data
radius = 23 m
angular acceleration = 5.7 rad/s²
to find out
time
solution
we know that radius is constant so that
tangential acceleration At = angular acceleration × radius ............. 1
tangential acceleration = 5.7 × 23 = 131.1 m/s²
and
radial acceleration Ar = (angular velocity)² × radius ........................2
we consider angular velocity = ω
this is acting toward center
so
compare 1 and 2
At = Ar
5.7 r =ω³ r
ω = √5.7 = 2.38746 rad/s
so
ω = 5.7 t
2.387 = 5.7 t
t = 2.387 / 5.7
t = 0.4187
time is 0.42 sec
The oscillation angular frequency of a drop with half of the first drop's radius is 4ω
<h3>What is surface tension?</h3>
Surface tension is the tension force exerted on an object by the surface of a liquid.
<h3>What is angular frequency?</h3>
Angular frequency is the frequency of oscillation of a rotating object. It is given in rad/s.
<h3>What is the oscillation angular frequency of a drop with half of the first drop's radius?</h3>
Given that
- the angular frequency of the drop is ω and
- radius r.
Since the energy of the drop is conserved, using the law of conservation of angular momentum, we have
Iω = I'ω' where
- I = initial rotational inertia of droplet = mr²
- where m = mass of drop and
- r = initial radius of droplet,
- ω = initial angular frequency of droplet,
- I' = initial rotational inertia of droplet = mr² where
- m = mass of drop and
- r' = final radius of droplet, and
- ω = final angular frequency of droplet
So, Iω = I'ω'
Making ω' subject of the formula, we have
ω' = Iω/I'
ω' = mr²ω/mr'²
ω' = r²ω/r'²
Given that the drop is half of the first drop's radius, r' = r/2
So, ω' = r²ω/r'²
ω' = r²ω/(r/2)²
ω' = r²ω/r²/4
ω' = 4ω
So, the oscillation angular frequency of a drop with half of the first drop's radius is 4ω
Learn more about angular frequency here:
brainly.com/question/28036464
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