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sergij07 [2.7K]
2 years ago
15

Bacteria vary in size, but a diameter of 2.0 µm is not unusual. What are the volume (in cubic centimeters) and surface area (in

square millimeters) of a spherical bacterium of that size?
Physics
1 answer:
telo118 [61]2 years ago
5 0

A = 4\pi r^2

A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2

A = 1.33*!0^{-5}MM^2

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In which direction does a bag at rest move when a force of 20 newtons is applied from the right? A. in the direction of the appl
-Dominant- [34]

The correct answer is A. In the direction of applied force. This is because acceleration occurs n the direction of applied force according to Newtons second law of motion which states that the acceleration of a body is directly proportional to the applied force and takes place in the direction of force.

5 0
3 years ago
Read 2 more answers
(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel
sammy [17]

Answer:

Part a)

R_1 = 0.072 m

Part b)

R_2 = 0.036 m

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

R = \frac{mv}{qB}

now for single ionized we have

R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}

R_1 = 0.072 m

Part b)

Similarly for doubly ionized ion we will have the same equation

R = \frac{mv}{qB}

R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}

R_2 = 0.036 m

Part c)

The distance between the two particles are half of the loop will be given as

d = 2(R_1 - R_2)

d = 2(0.072 - 0.036)

d = 0.072 m

6 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
If 478 watts of power are used in 14 seconds, how much work was done?
sergiy2304 [10]
" 478 watts " means 478 joules per second.

In 14 seconds, the total work or energy is (14 x 478) = <em>6,692 joules</em>
7 0
3 years ago
In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required
konstantin123 [22]

Answer:

2.46 eV

Explanation:

It is given that,

The energy of light that fall on the metal = 3.56 eV

The stopping potential of the metal = 1.1 V

We need to find the work function of the metal. It is given by the relation as follows :

W = E-KE ...(1)

Where KE is the kinetic energy of the ejected electron and it is given by :

KE = V×e

= 1.1 eV

Put all the values in formula (1)

W = 3.56 eV - 1.1 eV

= 2.46 eV

Hence, the work function of the metal is 2.46 eV. Hence, the correct option is (c).

5 0
2 years ago
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