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3 years ago

5

a magnetic flux of 4.72 x 10^-4 Wb passes through a loop oriented 65.5 degrees to a magnetic field of 1.44 T. what is the area o

f the loop

1 answer:

Digiron [165]3 years ago

7 0

Answer:7.904*10^-4

Explanation:

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Bacteria vary somewhat in size, but a diameter of 2.9 μm is not unusual.

Sedbober [7]

Answer:

a) 6.4 x 10^-12 cm^3

b) 17 x 10^-6 mm^2

Explanation

a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3

3

4

πr

3

V = \frac{4}{3}*3.142* 1.15^3

3

4

∗3.142∗1.15

3

= 6.3715 μm^3

1 μm^3 = 10^-12 cm^3

6.3715 μm^3 = 6.3715 x 10^-12 cm^3

==> 6.4 x 10^-12 cm^3

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2 years ago

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What wa can friction do to things like brake pads

andrey2020 [161]

Ware them down, its like rubbing two pieces of chalk together.

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3 years ago

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A manometer using oil (density 0.900 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases

Mila [183]

Answer:

Rise in level of fluid is 0.11 m

Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm

Solution:

As per the question:

Density of oil, $\rho_{o} = 0.900\ g/cm^{3} = 900\ kg/m^{3}$

Change in Pressure in the tank, $\Delta P = 7.28\ mmHg$

Density of the mercury, $\rho_{m} = 13.6\ g/cm^{3} = 13600\ kg/m^{3}$

Now,

To calculate the rise in the level of fluid inside the manometer:

We know that:

1 mmHg = 133.332 Pa

Thus

$\Delta P = 7.28\ times 133.332 = 970.656\ Pa$

Also,

$\Delta P = \rho_{o} gh$

where

g = acceleration due to gravity

h = height of the fluid level

$970.656 = 900\times 9.8\times h$

h = 0.11 m

Now, if mercury is used:

$\Delta P = \rho_{m} gh$

$970.656 = 13600\times 9.8\times h$

h = 0.00728 m = 7.28 mm

3 0

3 years ago

What happens if the volume of gas is reduced?

kolbaska11 [484]

Answer:

the volume will increase

Explanation:

if the temperature is held constant , the equation is reduced yo Boyle's law

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3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago