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3 years ago

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

Physics

2 answers:

AysviL [449]3 years ago

8 0

Answer:

Condensation causes dew to form on blades of grass. This is because the water molecules react to the grass when it is heated, and the surroundings became cool. The molecules would bring to calm and turn to liquid.

Explanation:

I just took the test

True [87]3 years ago

6 0

Answer:

AHAHHAH DEW DEW

Explanation:

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What is the energy of a photon with a frequency of 3. 6 × 1015 hz? planck’s constant is 6. 63 × 10–34 j•s.

GrogVix [38]

Answer:

2.029×10^-18 J

Explanation:

E=hv

E=(3.06×10^15)*(6.63×10^-34)

E=2.029×10^-18 J

3 0

3 years ago

Which of these processes in the water cycle occur at a very high rate, to cause a hurricane?

Pavel [41]

Answer: d. evaporation and condensation

Water vapor is known as the main fuel that moves the hurricane. The evaporation will cause the water vapor to move upward carrying the latent heat of condensation. The vapor will cause condensation later. If both happen at a very high rate, the wind produced can become a hurricane.

5 0

4 years ago

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

4 years ago

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

Rzqust [24]

Answer:

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

= 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0

3 years ago

011 10.0 points

Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

$v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s$