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2 years ago

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A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic

field at the (1 cm, 0 cm, 0 cm) position? Give your answer using unit vectors.

1 answer:

Sunny_sXe [5.5K]2 years ago

3 0

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

$B = \frac{\mu_0 qv sin\theta}{4\pi r^2}$

so here we know that for the direction of magnetic field we will use

$\hat B = \hat v \times \hat r$

so we have

$\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)$

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

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The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0 ( Gauge pressure)

Z₂= 0 m/s

$h_L=1.5\ m$

Now by putting the values

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

$Z_1-h_L=\dfrac{V_2^2}{2g}$

$3-1.5=\dfrac{V_2^2}{2\times 9.81}$

$V_2=\sqrt{2\times 1.5\times 9.81}\ m/s$

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0

3 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of

Flauer [41]

Answer:

$v = 4.375\,\frac{m}{s}$

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

$(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}$

The initial speed of Ryan is:

$v = 4.375\,\frac{m}{s}$

5 0

2 years ago

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

2 years ago

Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio

Pachacha [2.7K]

First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N

Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>

5 0

2 years ago

Read 2 more answers