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3 years ago

Which statement about these cells is correct

Physics

1 answer:

Nuetrik [128]3 years ago

5 0

B is correct because most of the cells look the same so it's B

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What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic field

goldenfox [79]

F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m

6 0

3 years ago

A 2.99-m-long2.99-m-long rod, as measured in its rest frame, speeds by you longitudinally at 6.49×107 m/s6.49×107 m/s . You meas

Rasek [7]

Answer:

The contraction in the rod is 71 mm.

Explanation:

Given that,

original length L'= 2.99 m

Speed $v= 6.49\times10^{7}\ m/s$

We need to calculate the length

Using expression for length contraction

$L'=\gamma L$

$L=\dfrac{L'}{\gamma}$

Where,

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$

$L=\sqrt{1-\dfrac{v^2}{c^2}}L'$

Where, v = speed of observer

c = speed of the light

Put the value into the formula

$L=\sqrt{1-\dfrac{(6.49\times10^{7})^2}{(3\times10^{8})^2}}\times2.99$

$L=2.919\ m$

The expression for the contraction in the rod

$d =L'-L$

$d=2.99-2.919$

$d=0.071$

$d= 71\ mm$

Hence, The contraction in the rod is 71 mm.

7 0

4 years ago

How to prove formula for volume of a sphere ?

atroni [7]

Answer:

The volume of radius is $\frac{4}{3}$ × π × radius³ Proved

Explanation:

Given as :

We know that volume of sphere is v = $\frac{4}{3}$ × π × radius³

Or, v = $\frac{4}{3}$ × π × r³

Let prove the volume of sphere

So, From the figure of sphere

At the height of z , there is shaded disk with radius x

Let Find the area of triangle with side x , z , r

<u>From Pythagorean theorem</u>

x² + z² = r²

Or, x² = r² - z²

Or, x = $\sqrt{r^{2}-z^{2} }$

Now, Area of shaded disk = Area = π × x²

Where x is the radius of disk

Or, Area of shaded disk = π × ( $\sqrt{r^{2}-z^{2} }$ ) ²

∴ Area of shaded disk = π × (r² - z²)

Again

<u>If we calculate the area of all horizontal disk, we can get the volume of sphere</u>

So, we simply integrate the area of all disk from - r to + r

i.e volume = $\int_{-r}^{r} \Pi(r^{2}-z^{2} )dz$

Or, v = $\int_{-r}^{r} \Pi r^{2}dz$ - $\int_{-r}^{r} \Pi z^{2}dz$

Or, v = π r² (r + r) - π $\frac{r^{3} -(-r)^{3})}{3}$

Or, v = π r² (r + r) - π $\frac{2r^{3}}{3}$

Or, v = 2πr³ - π $\frac{2r^{3}}{3}$

Or, v = 2πr³ ( $\frac{3-1}{3}$ )

Or, v = 2πr³ × $\frac{2}{3}$

∴ v = $\frac{4}{3}$ × π × r³

Hence, The volume of radius is $\frac{4}{3}$ × π × radius³ Proved . Answer

5 0

3 years ago

Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to complet

notsponge [240]

Answer:

The amount of energy required is $152.68\times 10^{3}Joules$

Explanation:

The energy required to convert the ice to steam is the sum of:

1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.

2) Latent heat required to convert the ice into water.

3) Energy required to raise the temperature of water from 0 degrees to 100 degrees

4) Latent heat required to convert the water at 100 degrees to steam.

The amount of energy required in each process is as under

1) $Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules$

where

$'S.heat_{ice}$ ' is specific heat of ice = $2.05J/^{o}C\cdot gm$

2) Amount of heat required in phase 2 equals

$Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules$

3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals

$Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules$

where

$'S.heat_{water}$ ' is specific heat of water= $4.186J/^{o}C\cdot gm$

4) Amount of heat required in phase 4 equals

$Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\$ $\\\\\\\\$ Thus the total heat required equals $Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules$