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4 years ago

What uniform magnetic field, applied perpendicular to a beam

Physics

1 answer:

schepotkina [342]4 years ago

5 0

Answer:

The magnetic field is $2.11\times10^{-5}\ T$

Explanation:

Given that,

Speed $v=1.30\times10^{6}\ m/s$

Radius = 0.350 m

We need to calculate the magnetic field

Using formula of magnetic field

$B =\dfrac{mv}{qr}$

Where, m = mass of electron

v = speed of electron

q = charge of electron

r = radius

Put the value into the formula

$B=\dfrac{9.1\times10^{-31}\times1.30\times10^{6}}{1.6\times10^{-19}\times0.350}$

$B=2.11\times10^{-5}\ T$

Hence, The magnetic field is $2.11\times10^{-5}\ T$

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3 years ago

Which of the following factors will negatively impact the rate of diffusion of a molecule?

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Answer:

A low difference in the concentration of the molecule across the media

Explanation:

Diffusion is a type of passive transport where the molecules move in the influence of concentration gradient of diffusing molecules i.e. from the higher concentration region to the lower concentration region. There are some factors which affect the rate of diffusion, these are written below -

Mass of diffusing molecule - lighter molecules diffuse faster and heavier one diffuse relatively slower.
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3 years ago

1.45 L of 16°C water is placed in a refrigerator. The refrigerator's motor must supply an extra 10.7 W power to chill the water

Vika [28.1K]

Answer:

The coefficient of performance of the refrigerator is 2.251.

Explanation:

In this case, the coefficient of performance of the refrigerator ( $COP$ ), no unit, is equal to the ratio of the heat rate received from the water to the power needed to work, that is:

$COP = \frac{\dot Q_{L}}{\dot W}$ (1)

$COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}$ (2)

Where:

$\dot Q_{L}$ - Heat rate received from the water, in watts.

$\dot W$ - Power, in watts.

$\rho$ - Density of water, in kilograms per cubic meter.

$V$ - Volume of water, in cubic meters.

$c_{w}$ - Specific heat of water, in joules per kilogram-degree Celsius.

$\Delta T$ - Temperature change, in degrees Celsius.

$\Delta t$ - Cooling time, in seconds.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 1.45\times 10^{-3}\,m^{3}$ , $c_{w} = 4187\,\frac{J}{kg\cdot ^{ \circ}C}$ , $\Delta T = 10\,^{\circ}C$ , $\dot W = 10.7\,W$ and $\Delta t = 2520\,s$ , then the coefficient of refrigeration of the refrigerator is: