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lawyer [7]
3 years ago
14

Which "spheres" are interacting when water evaporates from plants

Physics
2 answers:
Novay_Z [31]3 years ago
8 0
The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.
Volgvan3 years ago
4 0

Answer:

c

Explanation:

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A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from
o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}

There is NO initial vertical velocity, so:

\large\boxed{\Delta d= \frac{1}{2}at^2}}

Rearrange to solve for time:

2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}

6 0
3 years ago
A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°c to 125°
kotegsom [21]

Answer : The specific heat of aluminum is, 0.90J/g^oC

Solution : Given,

Heat absorbs  = 677 J

Mass of the substance = 10 g

Final temperature = 125^oC

Initial temperature = 50^oC

Formula used :

Q= m\times c\times \Delta T

or,

Q= m\times c\times (T_{final}-T_{initial})

Q = heat absorbs

m = mass of the substance

c = heat capacity of aluminium

T_{final} = final temperature

 T_{initial} = initial temperature

Now put all the given values in the above formula, we get the specific heat of aluminium.

677g= (10g)\times c\times (125-50)^oC

c=0.9026J/g^oC=0.90J/g^oC

Therefore, the specific heat of aluminum is, 0.90J/g^oC

7 0
3 years ago
Read 2 more answers
An arrow is projected by a bow vertically up with a velocity of 40 m/s, and reaches a target in 3 s. What is the velocity of the
Fittoniya [83]

Answer:

Explanation:

Step one:

given data

initial velocity u= 40m/s

time taken t=3seconds

final velocity v=?

Step two:

applying the first equation of motion

v=u-gt---  (the -ve sign implies that the arrow is against gravity)

assume g=9.81m/s^2

v=40-9.81*3

v=40-29.43

v=10.57m/s

Step three:

how high the target is located

applying

s=ut-1/2gt^2

s=40*3-1/2(9.81)*3^2

s=120-88.29/2

s=120-44.145

s=75.86m

6 0
3 years ago
100 Points !!!!!!!! I WILL GIVE BRAINLIEST
kumpel [21]

Answer: 1. 0.25 m/s squared

2. A = 0.4 m/s^2

3.=20kg. I did this before.

Explanation:

3 0
3 years ago
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A charge of 5.67 x 10^-18 C is placed 3.5 x 10^-6 m away from another charge of -
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+ 1.58 e -15

Please hit thanks button! :)
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