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Ilya [14]
3 years ago
11

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire ful

ly lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. The magnetic force on the wire is *

Physics
1 answer:
il63 [147K]3 years ago
8 0

Answer:

1.875*10^-5

Explanation:

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What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?
Marrrta [24]

Answer:

2.46\cdot 10^5 J

Explanation:

The enegy of a single photon is given by:

E=\frac{hc}{\lambda}

where

h is the Planck costant

c is the speed of light

\lambda is the wavelength of the photon

In this problem,

\lambda=486 nm=4.86\cdot 10^{-7}m

so the energy of one photon is

E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J

1 mole of photons contains a number of Avogadro of photons:

N_A = 6.022\cdot 10^{23}

therefore, the total energy of 1 mole of these photons will be

E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J

3 0
4 years ago
A bike at rest moves to 2 m/s in 2 seconds.
Sergio039 [100]

Answer:

Explanation:

When at rest, initial velocity = 0

In 2 seconds, final velocity = 2 m/s

Acceleration = (final velocity - initial velocity) / time

= (2 - 0) / 2

= 1 m/s^2

3 0
3 years ago
Read 2 more answers
You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ
Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

v^2-c^2=\dfrac{w^2}{t^2}

Put the value in the equation

6.00^2-c^2=\dfrac{w^2}{(20.1)^2}

36-c^2=\dfrac{w^2}{404.01}....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

v^2-c^2=\dfrac{w^2}{t^2}

Put the value in the equation

9.00^2-c^2=\dfrac{w^2}{(11.2)^2}

81-c^2=\dfrac{w^2}{125.44}....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}

45=w^2(0.00549)

w^2=\dfrac{45}{0.00549}

w=\sqrt{\dfrac{45}{0.00549}}

w=90.5

We need to calculate the current speed

Using equation (I)

36-c^2=\dfrac{(90.5)^2}{(20.1)^2}

36-c^2=20.27

c^2=20.27-36

c=\sqrt{15.73}

c=3.96\ m/s

(b). We need to calculate the shortest time

Using formula of time

t=\dfrac{d}{v}

t=\dfrac{90.5}{6}

t=15.0\ sec

We need to calculate the distance

Using formula of distance

d=vt

d=3.96\times15.0

d=59.4\ m

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

3 0
4 years ago
A particle travels in a circular orbit of radius 21 m. Its speed is changing at a rate of 23.1 m/s2 at an instant when its speed
Yuri [45]

The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components \vec a_c (c for center) and \vec a_t (t for tangent). Then its acceleration vector has magnitude

|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}

We have

\|\vec a_c\|=\dfrac{\|\vec v\|^2}r

where \|\vec v\| is the particle's speed and r is the radius of orbit, so

\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}

We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as \vec a_t, i.e. perpendicular to \vec a_c, so

\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}

Then the magnitude of the particle's acceleration is

\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}

7 0
3 years ago
An input force is a Push or pull that ________________________ to the machine
Jlenok [28]

Answer:

is applied to

Explanation:

An input force is push or pull that is applied to the machine and causes work to be done on it.

Input force for every machine should be lesser than the output force. The output force is the resulting force which produces work.

The force produced by any machine is a function of the input force and the design of the machine.

Any force applied to a machine is the input force.

8 0
4 years ago
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