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2 years ago

Explain why we have seasons on earth?

1 answer:

5 0

It’s because of the tilt of the earth’s axis, it means that w he n the earth leans towards the sun it’s summer, when it leans away from the sun it’s winter and in between the 2 spring and autumn (fall) occur.

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What is the force exerted on a charge of 2. 5 µC moving perpendicular through a magnetic field of 3. 0 × 102 T with a velocity o

stira [4]

The force acting on a moving charge is known as the magnetic force. The force acting on the charge will be 3.75 N.

<h3>What is the force exerted on the charge?</h3>

Magnetic fields only exert a force on a moving electric charge. A moving charge generates a magnetic field. With an increase in charge and magnetic field strength, this force rises.

when charges have higher velocities, the force is stronger. However, the magnetic force is always perpendicular to the velocity.

Mathematically the force exerted on the charge will be

F=qvBsinα

F= force acting on the charge

v = velocity of charge

q = charge

F=qvBsinα

F=2.5×10⁻⁶×5.0×10³×3.0×10²

F=37.5 N

Hence The force acting on the charge will be 3.75 N.

To learn more about the force acting on charge refer to ;

brainly.com/question/451411

F = q V B sinα

Where F is the force applied to a moving charge.

V = charge velocity

q stands for charge.

α = angle between V and B directions

As a result, the moving charge is subjected to a force of 3.75 Newton.

3 0

1 year ago

1. The term that describes where the supply curve intersects the demand curve is known as

nadezda [96]

Equilibrium is the answer

6 0

2 years ago

Total charge is uniformly distributed on a spherical surface of radius R. The sphere is centered at the origin and spins around

Flauer [41]

Answer:

Explanation:

i have a A in physics

5 0

2 years ago

Two forces F1 and F2 are acting on a block of mass m=1.5 kg. The magnitude of

nika2105 [10]

Answer:

Explanation:

Component of force F₁ in right direction = F₁cos37

= 12 x cos37 = 9.58 N .

Component of force F₂ in vertically upward direction = F₁sin37

= 12 x sin37 = 7.22 N .

a ) Let normal force be R

R + F₁sin37 = mg

R + 7.22 = 1.5 x 9.8 = 14.7

R = 7.48 N .

b )

Net force in horizontal direction = F₂ - F₁cos37

= F₂ - 9.58

This is equal to zero as body is moving with zero acceleration

F₂ - 9.58 = 0

F₂ = 9.58 N

c ) If body is moving with acceleration of 2.5 m /s² along the direction of F₂

F₂ - 9.58 = 1.5 x 2.5 = 3.75

F₂ = 13.33 N .

6 0

2 years ago

Saturated ethylene glycol at 1 atm is heated by a horizontal chromiumplated surface which has a diameter of 200 mm and is mainta

Paha777 [63]

Here is the full question.

Saturated ethylene glycol at 1 atm is heated by a chromium-plated surface which is circular in shape and has a diameter of 200-mm and is maintained at 480 K.

At 470 K the properties of the saturated liquid are mu = 0.38 * 10-3 N. s/m^2, Cp = 3280 J/kg. K and Pr = 8.7. The saturated vapour density is p= 1.66 kg/m^3. Take the liquid to surface constants to be Cnb = 0.010 and m=4.1.

Estimate the heating power requirement and the rate of evaporation

What fraction is the power requirement of the maximum power associated with the critical heat flux

Answer:

The heating power requirement = 559.2 W

The rate of evaporation = $6.89*10^{-4}kg/s$

The fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

Explanation:

From the thermodynamics tables; we deduced the value for enthalpy at the pressure 1 atm and $T_{sat}$ = 470 K for the saturated ethylene glycol.

Value for enthalpy of formation $h_{fg}$ = 812 kJ/kg

Density of saturated ethylene glycol $\rho___l$ = 1111 kg/m³

Surface tension $\sigma = 32.7*10^{-3}N/m$

The heat flux can be calculated by using the formula:

$q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2} [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3$

$= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )]$

= 308.56 × 576.6 × 0.1

= 1.78 × 10⁴ W/m²

Now; to find the heating power requirement; we have:

$q_{boil} = q__s }*A S$

= $1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2$

Thus, the heating power requirement = 559.2 W

The rate of evaporation is given as:

$m= \frac{q_{boil}}{h_[fg}}$

= $\frac{559.2}{812*10^3}$

= $6.89*10^{-4}kg/s$

Thus, the rate of evaporation = $6.89*10^{-4}kg/s$

To determine to what fraction in the power requirement of the maximum power is associated with the critical total flux ; we needed to first calculate the critical heat flux.

So, the calculation for the critical heat is given as: $q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}$

= $q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}$

= 200840.08 × 3.37

= 6.77 × 10⁵ W/m²

Finally, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is as follows:

= $\frac{q''s}{q''max}$

= $\frac{1.78*10^4}{6.77*10^5}$

= 0.026

Thus, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

7 0

3 years ago