Which of the processes of the water cycle occur by releasing energy

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

horsena [70]

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

3 0

3 years ago

The tension in the horizontal towrope pulling a water-skier is 250 N while the skier moves due west a distance of 50 m. How much

icang [17]

Answer:

W = 250(50) = 12500 J

Explanation:

5 0

3 years ago

Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential

mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0

3 years ago

An object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object. Use

shtirl [24]

Answer:

-30m/s

Explanation:

Given:

Initial velocity of object = 200 feet/second

Final velocity of object = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration "a" is given by:

$a=\frac{v_f-v_i}{t}$

where vf represents final velocity, vi represents initial velocity and is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a= -30m/s$

The acceleration of the object is -30m/s

3 0

3 years ago

A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density ρ is not un

baherus [9]

A negative charge and a positive charge

5 0

4 years ago