All categories

3 years ago

What is the magnetic force on a moving electric charge called

2 answers:

4 0

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule. The force is given by the charge times the vector product of velocity and magnetic field.

love history [14]3 years ago

4 0

Answer: i think The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule. The force is given by the charge times the vector product of velocity and magnetic field.

Explanation:

You might be interested in

Explain the differences among sand, silt, and clay, both in their physical characteristics and their behavior in relation to bui

3241004551 [841]

Answer:PHYSICALLY- sand is spherical in shape with large particles up to 0.18g

Silt also contains more of spherical particles with weights around 0.0029g.

Clay contain plate like particles,which are flat or layered. It's particles are generally lower that silt and sand below 0.0029g.

BEHAVIORS Sand particles are coarse and very porous,water can easily penetrate with no particles cohesion,does not retain water, difficult to expand,it is IDEAL FOR BUILDINGS.

Silt particles are also porous with some coarseness and no particles cohesion. retains water and can expand. NOT IDEAL FOR BUILDINGS.

Clay contain particles that are not coarse,they have high cohesiveness,they are not porous as it is difficult for water to penetrate. NOT IDEAL FOR BUILDINGS.

Explanation: Sand particles are coarse,very porous,contains spherical particles with large particles sizes and IDEAL FOR BUILDINGS

Silt particles are also porous, with some coarseness and with little or no particles cohesion, NOT IDEAL FOR BUILDINGS.

Clay soils are not porous water can not easily flow thought it,it is hard when dry and soft when wet. IT IS NOT A GOOD OPTION FOR BUILDINGS.

6 0

3 years ago

When the outside temperature is 5.2 ⁰C, a steel beam of cross-sectional area 52 cm2 is installed in a building with the ends of

il63 [147K]

Multiply the coefficient by the change in temperature:

1.1*10^-5 x (37-5.2) = 0.0003498

Multiply Young's modulus by the area by the above answer:

2*10^11 x 52 * 0.0003498 x (1/100)^2 = 3.63792 x 10^5 N

6 0

3 years ago

A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled

Vikentia [17]

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, $\mu =1.92\times 10^{- 3} Pa-s$

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

$F = \tau A$

where

$\tau$ = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = $(\tau ' +\tau '')A$

$\frac{F}{A} = \tau ' +\tau ''$

Also

F = $\frac{\mu v}{d}$

where

$\mu$ = dynamic coefficient of viscosity

Pressure, P = $\frac{F}{A}$

Therefore,

$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$

8 0

3 years ago

: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat

Snowcat [4.5K]

Answer:

Heat is lost at the rate of 750 J/s or W

The thermal resistance is 1 K/W

Explanation:

interior temperature $T_{2}$ = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature $T_{1}$ = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference $dt$ = $T_{2}$ - $T_{1}$ = 900 - 150 = 750 °C

note that $dt$ is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak $\frac{dt}{t}$