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Nataly [62]
4 years ago
10

A solar eclipse that occurs when the new moon is too far from Earth to completely cover the Sun can be either a partial solar ec

lipse or a(n)_____________.
Physics
1 answer:
liq [111]4 years ago
5 0

Answer:

ANULAR ECLIPSE

Explanation:

ANULAR ECLIPSE. Because the moon is very far, only a portion of the sun would be obscured, and then only the moon's outer ring will be viewable; this is called the anular eclipse.The ring of fire marks the maximum stage of an annular solar eclipse.

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A coil consists of 200 turns of wire. Each turn is a square of side d = 18 cm, and a uniform magnetic field directed perpendicul
vampirchik [111]

Answer:

Induced emf in the coil, \epsilon=4.05\ volts

Explanation:

Given that.

Number of turns in the coil, N = 200

Side of square, d = 18 cm = 0.18 m

The field changes linearly from 0 to 0.50 T in 0.80 s.

To find,

The magnitude of the induced emf in the coil while the field is changing.

Solution,

We know that due to change in the magnetic field, an emf gets induced in the coil. The formula of induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(NBA)}{dt}

A is the ares of square

\epsilon=AN\dfrac{d(B)}{dt}

\epsilon=AN\dfrac{B_f-B_i}{t}

\epsilon=(0.18)^2\times 200 \times \dfrac{0.5-0}{0.8}

\epsilon=4.05\ volts

So, the induced emf in the coil is 4.05 volts. Hence, this is the required solution.

8 0
3 years ago
A boy of mass, m weights 895 N. What is his mass, m?
Ksju [112]

Answer:

91.3 kg

Explanation:

weight = m*g

m: mass

g: gravity = 9.8 or 10 (depends on what your instructor tells you to use)

mass = w/g

895/9.8 = 91.3 KG

6 0
3 years ago
What is the length of the x-component of the vector shown below?
Mumz [18]

Answer: D. 2.6

Explanation:A P E X

7 0
3 years ago
Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei
mars1129 [50]

Explanation:

The given data is as follows.

   Fluid is water so, density \rho = 1000 kg/m^{3}

  Weight flow rate = 500 lbf/s = 2224.11 N/sec

  Cross-sectional area (A) = \pi \times \frac{30}{2} \times \frac{22}{2}

                                         = 0.05184 m^{2}

Hence, weight flow rate will be given as follows.

        w = \rho \times g \times A \times V

    2224.11 N/sec = \rho \times g \times A \times V

        V = \frac{2224.11}{1000 \times 9.81 \times 0.05184} m/s

            = 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

8 0
3 years ago
Read 2 more answers
Do you wont to be friends
Angelina_Jolie [31]

Answer:

shore

Explanation:

8 0
3 years ago
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