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4 years ago

15

_____ is a closed loop of conductors through which charges flow

1 answer:

Nataly_w [17]4 years ago

7 0

The answer is "b" a circuit.

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Which of these is the most likely prediction that the meteorologist would make about the weather? There is a possibility of

riadik2000 [5.3K]

Answer:

B. Dust storm

Explanation:

A meteorologist would predict dust storm most likely. Before dust storm the air condition and air pressure reduces in the particular area which indicate towards a dust storm. As pressure of a region reduces the air of the neighboring area try to maintain that pressure and rush towards low pressure area and causes dust storm.

6 0

3 years ago

Match the adaptations with their benefits.Match the adaptations with their benefits.

laila [671]

Wasp stinger protection from predators. Peacocks bright plumage to help obtain mate. Owls large eyes to help search for food. Bears fur to help maintain homeostasis

5 0

3 years ago

Read 2 more answers

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th

Katyanochek1 [597]

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf = - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

$=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV$

3 0

3 years ago

A brass statue with a mass of 0.40 kg and a density of 8.00×103kg/m3 is suspended from a string. When the statue is completely s

Mars2501 [29]

Answer:

ρ = 1469 kg/m³

Explanation:

given,

mass of statue = 0.4 Kg

density of statue = 8 x 10³ kg/m³

tension in the string = 3.2 N

density of the fluid = ?

Volume of the statue

$V = \dfrac{0.4}{8\times 10^3}$

V = 5 x 10⁻⁵ m³

W = ρ g V

W = ρ x 9.8 x 5 x 10⁻⁵

now, tension on the string will be equal to

T = mg - W

3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵

ρ x 9.8 x 5 x 10⁻⁵ = 0.72

ρ = 1469 kg/m³

8 0

3 years ago

If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amo

mestny [16]

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

$P = \frac{V^2}{R}$

Power lost in the first wire;

$P_1 = \frac{V_1^2}{R}$

$R = \frac{V_1^2}{P_1}$

Power lost in the second wire

$P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}$

Keeping the resistance constant, we will have the following equation;

$\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\$

$P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1$

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

8 0

3 years ago