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3 years ago

14

17.

1 answer:

avanturin [10]3 years ago

6 0

Answer:

Gamma rays

Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies.

Explanation:

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Which of these is a mixture?

kicyunya [14]

It is salt water because they are two things combined together.

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3 years ago

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A circuit has a voltage drop of 60.0 V across an 80.0 resistor that carries a current of 0.750 A. What is the power used by the

ira [324]

45.0 W is the answer

4 0

3 years ago

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

$\rho=\frac{m}{V}$

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

7 0

3 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

3 years ago

Calculate the velocity of the objects motion represented in the following graph. Show all your work.

Leni [432]

as we know that velocity= displacement/change in time

hence velocity here is= 11-3/12-0

=8/12

=2/3

=0.667m/s

hopefully i am right

6 0

3 years ago