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Mrrafil [7]
2 years ago
8

PLEASE HELP ASAP BEST ANSWER WILL BE MARKED BRAINLIEST!!!!!

Physics
1 answer:
umka2103 [35]2 years ago
3 0

Answer:

i think its The movement of large pieces of ice from one place to another.

Explanation:

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A _____ is used in a motor to switch the direction of the magnetic field created by the current.
insens350 [35]

Answer:

A _commutator_ is used in a motor to switch the direction of the magnetic field created by the current.

The rotating part of a motor that holds the electromagnets is called the __armature___.

Electric current passes through the _brushes_ and into the electromagnets in an electric motor.

A motor turns _electrical_ energy into _mechanical_ energy.

Explanation:

A commutator, which is a split ring rotary switching device, reverses the direction of the current between the external circuit and the rotor. Reversing the current reverses the magnetic field.

The armature comprises the rotating part of the motor and the electromagnets

A brush is the electrical contact for conducting current through the moving and stationary parts of an electric motor

An electric motor turns electrical energy into mechanical energy.

8 0
3 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra
Norma-Jean [14]

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

4 0
3 years ago
What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze
yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of 62.5 cm of H_2O

i.e. h=62.5 cm =0.625 m

Effective area A=51 cm^2

initial Pressure= 1 atm=101.325 kPa

Gauge Pressure P=\rho gh

\rho =density\ of\ water =1000 kg/m^3

P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa

Force creates a pressure of P_1 which will be equal to Gauge Pressure

P_1=\frac{F}{A}

P_1=P_{gauge}

\frac{F}{A}=5.937 kPa

F=5.937\times 51\times 10^{-4}\times 10^3

F=30.27 N

6 0
3 years ago
What is 9414 divided by 18​
Levart [38]
9414/18 is equal to 523
3 0
3 years ago
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