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seropon [69]
3 years ago
15

Explain the process that causes dew to form on blades of grass?

Physics
2 answers:
luda_lava [24]3 years ago
7 0
<span>Heat is radiated, atmospheric moisture condenses at a rate greater than that at which it can evaporate, resulting in the formation of water droplets.</span>
Varvara68 [4.7K]3 years ago
4 0
Condensation causes dew to form on blades of grass because the water molecules that were in the air were excited because they were heated, and when the surroundings became cooler, the molecules calmed down and condensed into liquid.
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A car travels from 20-meters to 60-meters in 10 seconds. Calculate the car's speed.
slamgirl [31]

Answer:

8 m/s

Explanation:

All you have to do here is add 20 and 60 (giving you 80) and dividing by 10 seconds. 80/10= 8 m/s

3 0
3 years ago
good morning to you all! can someone please answer this, ill give you brainliest and your earning 50 points.
elena55 [62]

Despite the risks, Human populations can be found near volcanoes. This can be associated with the availability of which natural resource?

  • <em>Geothermal</em><em> </em><em>energy</em>

<u>Geothermal</u><u> </u><u>energy</u><u> </u><u>is</u><u> </u><u>the</u><u> </u><u>f</u><u>o</u><u>r</u><u>m</u><u> </u><u>of</u><u> </u><u>energy</u><u> </u><u>obtained</u><u> </u><u>from</u><u> </u><u>the</u><u> </u><u>heat</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>earth</u><u>.</u><u> </u><u>Near</u><u> </u><u>volcanoes</u><u> </u><u>the</u><u> </u><u>land</u><u> </u><u>is</u><u> </u><u>more</u><u> </u><u>hot</u><u> </u><u>compared</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>normal</u><u> </u><u>areas</u><u>.</u><u> </u><u>So</u><u>,</u><u> </u><u>the</u><u> </u><u>heat</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>lava</u><u> </u><u>will</u><u> </u><u>help</u><u> </u><u>in</u><u> </u><u>producing</u><u> </u><u>Geothermal</u><u> </u><u>energy</u><u>.</u><u>.</u><u>.</u><u>~</u>

7 0
2 years ago
Read 2 more answers
The following equation, N2 + 3 H2 —&gt;2 NH3 ,describes a
mafiozo [28]
Physical change 1 is the answer
4 0
2 years ago
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A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th
zhannawk [14.2K]

Answer: V_{f}=2.96m/s    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

In the X-component:

\sum F_{x}=m.a

m.a=w_{x}    (5)

Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

d=\frac{1}{2}a{t}^{2}   (10)

We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

V_{f}=a.t   (14)

V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

Finally we get the value of the Final Velocity of the block:

V_{f}=2.96m/s    

6 0
3 years ago
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