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postnew [5]
3 years ago
10

- Calculate the efficiency of the human body if a person eats a hamburger containing 7.4822 kJ of

Physics
2 answers:
rewona [7]3 years ago
5 0
W = E
efficiency = (Woutput /Winput) x 100%= Eo/Ei x100%
= (2021 j / 7.4822 kj) x 100%
= 27.01% == B
levacccp [35]3 years ago
3 0

Explanation:

w = E

efficiency = (Woutput /Winput) x 100%= Eo/Ei x100%

= (2021 j / 7.4822 kj) x 100%

= 27.01% == B

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Viktor [21]
It would be A because it would make sense
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A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

\phi = 1.89 \times 10^{-7}t

(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0
3 years ago
A uniform horizontal rod of mass Mand length rotates with angular velocity about a vertical axis through its center. Attached to
Lostsunrise [7]

Answer:

     o

90

Explanation:

3 0
3 years ago
A. 36 mins 52.5 secs<br> b. 24mins 22.5 secs<br> c. 13 mins 0 secs<br> d. 9 mins 45 secs
DedPeter [7]
I think it might be A. I’m sorry if I’m wrong
5 0
3 years ago
Read 2 more answers
We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in
leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

B=\mu_0\, \frac{N}{L} I

Then, if we assign the subindex "1" to the quantities that define the magnetic field (B_1) inside solenoid 1, we have:

B_1=\mu_0\, \frac{N_1}{L_1} I_1

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then L_2=L_1/2

2) twice as many turns as solenoid 1. Then N_2=2\,N_1

3) three times the current of solenoid 1. Then I_2=3\,I_1

we obtain:

B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1

5 0
3 years ago
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