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3 years ago

6

A printer is connected to a 1.0 m cable. if the magnetic force is 9.1 × x10-5 n, and the magnetic field is 1.3 × 10-4 t, what is

the current in the cable

1 answer:

Valentin [98]3 years ago

5 0

The magnetic force on a current-carrying wire due to a magnetic field is given by
$F=ILB$
where
I is the current
L the wire length
B the magnetic field strength

In our problem, L=1.0 m, $F=9.1 \cdot 10^{-5} N$ and $B=1.3 \cdot 10^{-4}T$ , so we can re-arrange the formula to find the current in the wire:
$I= \frac{F}{LB}= \frac{9.1 \cdot 10^{-5} N}{(1.0 m)(1.3 \cdot 10^{-4} T)} =0.7 A$

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•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0

Firdavs [7]

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+ (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

7 0

3 years ago

microwave ovens rotate at a rate of about 6.3 rev/min. what is this in revolutions per second? (you do not need to enter any uni

levacccp [35]

The microwave ovens rotate at a rate of about 0.105 rev/s.

The microwave rotation is the number of revolutions in a unit of time. To change the unit for angular velocity, assume that the quantity is multiplied by the unit it has. Then change to the desired units. The angular velocity is denoted by ω and has a magnitude of 6.3 rev/min.

ω = 6.3 rev/min

$\omega = 6.3 \times \frac{1 \: rev}{1 \: min}$

1 minute = 60 seconds
The revolution unit didn't change

$\omega = 6.3 \times \frac{1 \: rev}{60 \: seconds}$

$\omega = \frac{6.3 \: rev}{60 \: seconds}$

ω = 0.105 rev/s

Learn more about Angular velocity here: brainly.com/question/29344944

#SPJ4

7 0

1 year ago

Who was the first person referred to as a psychologist?

musickatia [10]

Answer:

Wilhelm Wundt

Explanation:

3 0

4 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago

Light of wavelength 530.00 nm is incident normally on a diffraction grating, and the first‑order maximum is observed to be 33.0∘

GarryVolchara [31]

Answer:

1028 slits/mm

Explanation:

We are given that

Wavelength of light, $\lambda=530nm=530\times 10^{-9} m$

1nm= $10^{-9} m$

$\theta=33^{\circ}$

n=1

We have to find the number of slits per mm are marked on the grating.

We know that

$dsin\theta=n\lambda$

Using the formula

$dsin33^{\circ}=1\times 530\times 10^{-9}$

$d=\frac{530\times 10^{-9}}{sin33^{\circ}}$

$d=9.731\times 10^{-7} m$

1m= $10^{3}mm$

$d=9.731\times 10^{-7}\times 10^3$ mm

$d=0.0009731mm$

Number of slits= $\frac{1}{d}$

Number of slits= $\frac{1}{0.0009731}$ /mm

Number of slits=1028/mm

Hence, 1028 slits/mm are marked on the grating.

8 0

3 years ago

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