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Tasya [4]
3 years ago
7

Angular velocity and linear velocity have the same dimensions. This statement is:

Physics
1 answer:
iragen [17]3 years ago
5 0
I think it would be C: Sometimes true and sometimes false. It varies.

Hope this helped, have a great day! :D
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A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc
hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

inserting all the values

\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

      N₂² = 4

      N₂ = 2 mm

hence , the correct answer is option E

4 0
3 years ago
The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a
Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ \frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2}  }{Distance_{old}^{2}  }

Now, we know the new distance is half the original distance,

F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}

F_{new} = 4F_{old}

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

#SPJ4

8 0
1 year ago
A proton traveling due north enters a region that contains both a magnetic field and an electric field. The electric field lines
Bess [88]

Answer:

upward

Explanation:

In the electromagnetic system of force if the direction of motion of proton does not changes it means that the electric and magnetic forces are such a ways that they are cancelling each other's effect.

Since, electric field lines will exert a force on the proton to the west, hence, the magnetic force must force it to the east. It is well known that magnetic force acts perpendicular to the direction of magnetic field.  magnetic field should point upward direction.

3 0
3 years ago
Question 4 of 10
Alexus [3.1K]

Answer:

22.3 kg•m/s

Explanation:

Apex;)

7 0
3 years ago
A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu
kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the​ rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the​ cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) =  ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i)  work required to pump the contents to the​ rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the​ cone's rim

Extra work required = 28,914.229 lb ×  5ft = 144571.145 lb.ft

Total work required = (281,913.733 +  144571.145) lb.ft

                                 = 426,484.878 lb.ft

5 0
3 years ago
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