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3 years ago

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Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as muc

h as Jack's power output. four times Jack's power output. twice Jack's power output. one-half as much as Jack's power output. the same as Jack's power output.

1 answer:

Musya8 [376]3 years ago

3 0

Answer:

Second Choice.

Explanation:

Jack's Power = W/t

Jill's Power = 2W/(0.5)*t

2/0.5 = 4

Jill's Power = 4*W/t

Jill's Power is 4 times greater than Jack's

Second Choice

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In a transformer, energy is carried from the primary coil to the secondary coil by:________

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In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

To find the answer, we have to know more about the transformer.

<h3>How transformer works?</h3>

An item utilized in the transfer of electric energy is a transformer.
AC current is used for transmission.
It is frequently used to modify the supply voltage between circuits without altering the AC frequency.
The fundamentals of mutual and electromagnetic induction govern how the transformer operates.
Magnetic field through the primary coil changes when primary coil current varies. the iron core of the secondary coil likewise has a magnetic field.
EMF is therefore generated in the secondary coil.

Thus, we can conclude that, in a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

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2 years ago

Find the magnitude of the sum<br> of these two vectors:<br><br> 101 m<br> 60.0 °<br> 85.0 m

attashe74 [19]

Answer: 161.3

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3 years ago

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

4 years ago

A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward

Anni [7]

Answer:

$m=18000kg$

Explanation:

From the question we are told that:

Crane Length $l=20m$

Crane Mass $m_a=3000kg$

Arm extension at lifting end $l_l=15m$

Arm extension at counter weight end $l_c=5m$

Load $M_l=5000kg$

Generally the equation for Torque Balance is mathematically given by

$T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0$

$mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0$

$m=18000kg$

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3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

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3 years ago