Recall: earth applies _____________ on Earth or equivalently

A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of

serg [7]

Answer:

Explanation:

vertical magnetic field B_v = 4 x 10⁻⁵ T.

Magnetic field due to horizontal current at point 20 cm above

= (μ₀/4π ) x (2i / R)

= 10⁻⁷ x 2 x 20/ 20 x 10⁻²

= 2 x 10⁻⁵ T

It will act coming out of paper. Hence it will be normal to magnetic field given .

So resultant magnetic field

= √ (4² + 2²) x 10⁻⁵ T

= 4.47 X 10⁻⁵ T .

8 0

3 years ago

In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.00 × 10−14 m from a fragment that has 46 proto

Valentin [98]

Answer:

electric potential is 3.31 × $10^{6}$ V

potential energy is 152 MeV

Explanation:

given data

fragment charge Q = 46 protons = 46 × 1.6 × $10^{-19}$ C

to find out

electric potential and potential energy

solution

we know here distance from fragment d = 2 × $10^{-14}$ m

and constant for electric force k that is 9 × $10^{9}$ N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 × $10^{9}[/tex ×46 × 1.6 × [tex]10^{-19}$ / ( 2 × $10^{-14}$ )

electric potential = 3.31 × $10^{6}$ V

and

we know relation between electric potential and potential

that is V = U/q

so U will be = qV

now put all value

we get potential energy U

potential energy = 46 × 3.31 × $10^{6}$

potential energy = 1.52 × $10^{8}$ eV

so potential energy = 152 MeV

4 0

3 years ago

A 49.5-turn circular coil of radius 5.10 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

ipn [44]

Answer:

The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm

Explanation:

Given;

number of turns of the circular coil, N = 49.5 turns

radius of the coil, r = 5.10 cm = 0.051 m

magnitude of the magnetic field, B = 0.535 T

current in the coil, I = 26.5 mA = 0.0265 A

The magnitude of the maximum possible torque exerted on the coil is calculated as;

τ = NIAB

where;

A is the area of the coil

A = πr² = π(0.051)² = 0.00817 m²

Substitute the given values and solve for the maximum torque

τ = (49.5) x (0.0265) x (0.00817) x (0.535)

τ = 0.00573 Nm

τ = 5.73 x 10⁻³ Nm

3 0

3 years ago

A ball resting on a roof 75 meters high has 1000 Joules of gravitational potential energy. Calculate the mass of the ball. (SHOW

Arturiano [62]

Answer:

The mass of the ball is 1.360 kilograms.

Explanation:

By Work-Energy Theorem, gravitational potential energy ( $U$ ), in joules, is the product of weight of the ball ( $W$ ), in newtons, and height ( $h$ ), in meters. Please notice that weight is the product of the mass of the ball ( $m$ ) and gravitational acceleration ( $g$ ), in meters per square second. Then, the formula for the mass of the ball is:

$m = \frac{U}{g\cdot h}$ (1)

If we know that $U = 1000\,J$ , $g = 9.807\,\frac{m}{s^{2}}$ and $h = 75\,m$ , then the mass of the ball is:

$m = \frac{U}{g\cdot h}$

$m = 1.360\,kg$

The mass of the ball is 1.360 kilograms.

8 0

3 years ago

Carol is farsighted ( presbyopia) and cannot see objects clearly that are closer to her eyes than about meter. She sees objects

egoroff_w [7]

Answer:

1.0 dioptres

Explanation:

Farsightedness is an eye defect in which a person can see far objects clearly but not near objects. That implies that the patients' near point is farther than 25cm which is the normal least distance of distinct vision.

Farsightedness results from the eyeball being too long or the crystalline lens not being sufficiently converging.

Carol is farsighted with a near point of about a meter (100cm). We desire to make a lens to enable her near point be reduced to about 50cm. The focal length and power of this lens is calculated in the image attached.

The power of a lens is the inverse of its focal length in meters hence the 100 in the formula for power of the lens.

7 0

3 years ago

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