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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,

Physics
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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Both independent and dependent clauses
Assoli18 [71]

Answer: have a subject and a verb

Explanation:

5 0
3 years ago
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Rock at the top of a 20 m tall hill the rock has a mass of 10 kg how much potential energy does it have
Blizzard [7]
PE = 10 * 10 * 20 = 2000 Joule
6 0
3 years ago
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The scale is 1:30 and their are 10 boulders. How many boulders are their in the real pond
Alenkasestr [34]

Answer:

the correct answer is C

Explanation:

When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.

Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.

Applying this to our case if there is

               10 boulder on the chart

in reality there are #_boulder = 10 30

                                #_boulder = 300 boulder

so the correct answer is C

7 0
3 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t
zlopas [31]

Answer:

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

Explanation:

As we know that the wave equation is given as

y = A sin(\omega t - k x + \phi_0)

now we have

A = 0.19 m

\lambda = 2.6 m

so we have

k = \frac{2\pi}{\lambda}

k = \frac{2\pi}{2.6}

k = 2.42  per m

also we have

T = 1.2 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{1.2}

\omega = 5.23 rad/s

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

\phi_0 = \frac{\pi}{2}

so we have

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

6 0
3 years ago
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