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Alenkasestr [34]
3 years ago
6

What do you expect the enclosed current is for a bar magnet?

Physics
1 answer:
BartSMP [9]3 years ago
6 0
The point P lies at a linear distance 1 m from the corner. Magnetic field due to current in wire Magnetic field due to current in wire. Applying Right hand thumb rule, the magnetic field at P is perpendicular to xy plane and into the plane of drawing (i.e. negative z-direction).
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Who the queen of rap
Inessa [10]

Answer:

Nicki or cardi b idek....

6 0
2 years ago
Read 2 more answers
Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of
FrozenT [24]
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
6 0
3 years ago
A cyclist maintains a constant velocity of 6.1 m/s headed away from point A. At some initial time, the cyclist is 242 m from poi
KengaRu [80]

Answer:

553.1m

Explanation:

When an object moves at constant velocity we can express this movement like V=x/t, where V is the velocity, x is the displacement and t is the time spent on it.

In that way, the expression x=V.t give us the displacement from t=0s until t=51s, but we have to sum the initial distance from the point A.

x=242m +V.t = 242m + (6.1m/s x 51s) = 553.1m

7 0
3 years ago
A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th
Hitman42 [59]

Answer:

Explanation:

Given

inclination \theta =30^{\circ}

initial speed u=20\ m/s

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

t_1=\frac{2u\sin \theta }{g}

t_1=\frac{2\times 20\times \sin 30}{10}

t_1=2\ s

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

h=u_yt+\frac{1}{2}a_yt^2

where, h=height

u_y=vertical velocity

a_y=vertical acceleration

t_0=time

45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2

t_0^2=\frac{70}{9.8}

t_0=2.64\ s

thus total time time required is t=t_0+t_1=2.64+2=4.64\ s

vertical velocity just before hitting

v_y=\sqrt{u_y^2+2\times a_y\times s}

v_y=\sqrt{10^2+2\times 10\times 45}

v_y=\sqrt{1000}=31.622\ m/s

Horizontal velocity v_x=u\cos 30=17.32\ m/s

Net velocity Just before hitting =\sqrt{v_x^2+v_y^2}

=\sqrt{(17.32)^2+(31.62)^2}

=\sqrt{1299.82}=36.05\ m/s

                 

7 0
3 years ago
A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700 m2 .
Darina [25.2K]

Answer:

The  value is  E =  8.9 *10^{-5} \  J

Explanation:

From the question we are told that

   The  area is  A =  0.700 \  m^2

   The  root mean square value  is E_{rms} =  0.0400 \  V/m

   The time taken is t =  30.0 \  s

Generally the energy is mathematically represented as

     E =  c *  \sepsilon_o *  A  *  t  * E_{rms}^2

=>    E =  3.0*10^{8} *  8.85*10^{-12} *  0.700 *  30 * (0.04)^2

=>     E =  8.9 *10^{-5} \  J

     

3 0
3 years ago
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