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jasenka [17]
3 years ago
6

What can Lisa do to increase the strength of the electromagnet? She can use a nail with weaker magnetic properties. She can chan

ge the direction of the nail. She can increase the number of wire loops. She can reduce the current in the wire.
Physics
2 answers:
Kryger [21]3 years ago
7 0

Answer:

C

Explanation:

Edge2021

Illusion [34]3 years ago
3 0

Answer:She can increase the number of wire loops

Explanation: When the number of loops of wire ( usually around the iron ) increases, the strength of the electromagnet is increased.

According to the equation

B=μ₀ N I

Where

μ₀ = permeability of the core

N= Number of turns of the coil

I= Current flowing through the coil

From the equation above , we can see that The strength of an electromagnet depends the amount of current, no of turns of coil and the permeability  core of coil. we can now say that Increasing the  number of wire loops  increases the strength of electromagnet.

Therefore, Lisa should increase the number of wire loops to increase the strength of the electromagnet

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Can someone list 6 advantages of renewable energy.
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Answer:

Solar energy,  Wind energy,  Hydro energy, Tidal energy, Geothermal energy,  Biomass Energy.

Explanation:

I hope that it helps you...

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3 years ago
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5 0
3 years ago
Read 2 more answers
What is the speed of a wave that has a wavelength of 225 m and a frequency of 117 hz ? Show your work
avanturin [10]

Answer:

26325 m\s

Explanation:

Data:

v = ?

f = 117 Hz

w = 225

Formula:

v = fw

Solution:

v = ( 117)(225)

v = 26325 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

6 0
3 years ago
An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what
Katarina [22]

Answer:

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3 0
2 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
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