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LuckyWell [14K]
4 years ago
10

What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so

the current drops to essentially zero in 0.015 seconds
Physics
1 answer:
puteri [66]4 years ago
7 0

Answer:

93 V

Explanation:

use the expression for induced emf as

EMF = L dI/dt

L = 560 x 10⁻³

dI = 2.5 A  

dt = 0.015 s = 15 x 10⁻³ s

EMF = L dI/dt

= 0.560 * 2.5/0.015

= 93.33 V

≅ 93 V

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A ball is rolled of a 2.14m table at 8m/s. Find the time of flight for the ball and the horizontal range
Mama L [17]

Answer:

0.661 s, 5.29 m

Explanation:

In the y direction:

Δy = 2.14 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.661 s

In the x direction:

v₀ = 8 m/s

a = 0 m/s²

t = 0.661 s

Find: Δx

Δx = v₀ t + ½ at²

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Round as needed.

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Olympic gold medalist Michael Johnson runs one time around the track 400 meters in 38 seconds what is his displacement what is h
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Use the facts below to answer the question: Who reaches One-Eyed Willie’s pirate ship first?
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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
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