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3 years ago

Which subatomic particle has a negative charge?

Physics

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

C. $\displaystyle electron$

Explanation:

0 charge → Neutron

1 charge → Proton

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A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the

Phantasy [73]

Answer:

3.4 x 10^-4 T

Explanation:

A = 1.5 x 10^-3 m^2

N = 50

R = 180 ohm

q = 9.3 x 106-5 c

Let B be the magnetic field.

Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .

Let e be the induced emf and i be the induced current

e = rate of change of magnetic flux

e = dФ / dt

i / R = B x A / t

i x t / ( A x R) = B

B = q / ( A x R)

B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T

6 0

4 years ago

A 1-ton shark swims to the right toward a beach. In which direction is the reaction force to the shark's weight?

Natasha_Volkova [10]

The Answer you are looking for is D.

7 0

3 years ago

A sound system is designed to produce a 1.0W/m2 sound intensity over the surface of a hemisphere 20 m in radius. What acoustic p

Marina CMI [18]

Answer:

2513.6 W

Explanation:

Acoustic power = sound intensity × area of hemisphere

Sound intensity = 1 W/m^2

Area of hemisphere = 2πr^2 = 2×3.142×20^2 = 2513.6 m^2

Acoustic power = 1 W/m^2 × 2513.6 m^2 = 2513.6 W

3 0

4 years ago

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

Since the speed of sound is not given in the question, we would assume it to be 340m/s

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

So, the length of the second pipe is L – L’

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

The beat frequency due to the fundamental frequencies of the first and second pipe is

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago

Scientists in the laboratory create a uniform electric field e⃗ = 6. 0×105 k^v/m in a region of space where b⃗ =0⃗

zvonat [6]

Scientist in the laboratory create uniform electric field.

PART A: The electric field in rocket frame is (0,0, 6.0 × 10 ⁵).

PART B: The magnetic field in the rocket frame is 7.3 × 10⁻⁶ j∧ .

<h3>What is uniform electric field?</h3>

The electric field whose strength is not changing with respect to time is called uniform magnetic field.

The electric field is 6.0 × 10 ⁵ k^v/m and the magnetic field is zero.

E' = E + (V×B)

= 6.0 × 10 ⁵ + (1.1 × 10⁶ × 0)

E' = 6.0 × 10 ⁵ k∧

(Ex , Ey, Ez) = (0,0, 6.0 × 10 ⁵)

The new magnetic field is represented as

B' = B - 1 /c²(v × E)

=0 - 1 / (3× 10⁸)²(1.1 × 10⁶ i∧ × 6.0 × 10 ⁵k∧ )

B' = 7.3 × 10⁻⁶ j∧

PART A: The electric field in rocket frame is (0,0, 6.0 × 10 ⁵).

PART B: The magnetic field in the rocket frame is 7.3 × 10⁻⁶ j∧ .

Learn more about uniform electric field.

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7 0

2 years ago