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Citrus2011 [14]

3 years ago

12

4. Some early mornings , dew drops can be found on grass or a car parked outside,

1 answer:

horsena [70]3 years ago

3 0

In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level

Dew can be formed on any object when the temperature of the object drop. When this happen, the object will be cool which will eventually cool the surrounding air around the object.

Dew drops is as a result of condensation in the air. When the cool air causes the air vapor to convert to liquid. The dew will form when the temperature of the object balances with the dew point in the surrounding environment.

In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level

Therefore the correct option is therefore A

Learn more here : brainly.com/question/13834972

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What two things do you need to know to describe the velocity of an object?

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Hey scooters dragging of 520 kg walk-through forest at a constant speed of 3.5 m/s. If the scooter is applying a force of 1850 N

Simora [160]

The coefficient of friction is 0.363

Explanation:

There are two forces acting on the scooter in the horizontal direction:

- The applied force, F = 1850 N, forward

- The frictional force, $F_f$ , backward

Since the scooter is moving at constant speed, the acceleration is zero, so the net force acting on the scooter must be zero. Therefore we can write:

$F-F_f = 0\\F_f = F = 1850 N$

The frictional force can be written as

$F_f = \mu R$ (1)

where

$\mu$ is the coefficient of friction

R is the normal reaction of the ground on the scooter

For a flat horizontal surface, there is equilibrium along the vertical direction, so the normal reaction is equal to the weight:

$R = W = mg$

where

m = 520 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting into (1),

$F_f = \mu mg = 1850 N$

and solving for $\mu$ ,

$\mu=\frac{F_f}{mg}=\frac{1850}{(520)(9.8)}=0.363$

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3 years ago

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A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of

qaws [65]

Answer:

$\frac{E_1}{E_2}= 4$

Explanation:

Capacitance C is given by

$C= \frac{\epsilon_0A}{d}$

A= area of capacitor cross section

d= distance

therefore,

$C_1= \frac{\epsilon_0A_1}{d_1}$

A_1= πR^2

d_1= d

$C_2= \frac{\epsilon_0A_2}{d_2}$

A_= π(2R)^2

d_2 = 2d

$q= \frac{\epsilon_0A_1}{d_1}V_1$

threfore

$V_1= \frac{qd_1}{\epsilon_0A_1}$

and

$V_2= \frac{qd_2}{\epsilon_0A_2}$

also we know that E= V/d

⇒ $\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}$

⇒ $\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}$

= A_1/A_2= $\frac{4R^2}{R^2}$ =4

therefore,

$\frac{E_1}{E_2}= 4$

5 0

3 years ago