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1 year ago

What is the term for producing a current by moving a wire through a magnetic field?.

Physics

1 answer:

maria [59]1 year ago

6 0

The answer is electromagnetic Induction.

I hope this answer will help you

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A motorcycle has a speed of 30 m / s. After braking, it decelerates with constant deceleration A = -3.0 m / s ^ 2

marin [14]

$a = \displaystyle\frac{v-u}{t}$

$-3 = \displaystyle\frac{v - 30}{3}$

$v -30 = -9$

$v = 21 m/s$

4 0

3 years ago

Read 2 more answers

1) According to the law of conservation of energy, A) an object loses most of its energy as friction. B) the total amount of ene

Blababa [14]

Your answer is C) <span> the potential energy of an object is always greater than its kinetic energy </span>

5 0

3 years ago

A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str

ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0

3 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

2 years ago

Can someone help me?!!!!!

ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

$(-3\cdot (-6), -2 \cdot (-6))=(18,12)$

The magnitude is given by Pythagorean's theorem:

$m=\sqrt{18^2+12^2}=21.6$

and the direction is given by the arctan of the ratio between the y-component and the x-component:

$\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}$

3 0

3 years ago