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Anton [14]
2 years ago
8

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

find the magnitude and direction of the force this field exerts on a charge of +6.80 μc.
Physics
1 answer:
denis23 [38]2 years ago
5 0
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
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A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk
Sonbull [250]

To solve this problem we will apply the concepts related to the centripetal force, for which it is necessary to equate it with the static friction force of the body. From this, we will clear the speed and replace with the given values. Our values are defined as,

r = 16m

m = 82kg

\mu_s = 0.63

Maximum velocity can be find out using centripetal force,

F_c = \frac{mv^2}{r}

Must be equal to,

\frac{mv^2}{r} = \mu_s mg

v = \sqrt{\mu_s gr}

v = \sqrt{(0.63)(9.8)(16)}

v = 9.93m/s

Therefore the maximum speed that he can travel through the arc without slipping is 9.93m/s

3 0
2 years ago
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards
8_murik_8 [283]

Answer:

<em>b. The current in the loop always flows in a counterclockwise direction.</em>

<em></em>

Explanation:

When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.

3 0
2 years ago
calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame
kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0
2 years ago
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Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.

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Here h is the Planck's constant,c is the velocity of light.

Here we have been given red light and blue light.

In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.

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<span>PA = 1.06 atm and PB = 0.53 atm</span>
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