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krok68 [10]
3 years ago
12

Cutting down trees can decrease the amount of water vapor entering the air through the processes of ____?

Physics
1 answer:
AlexFokin [52]3 years ago
5 0

✿  Transpiration is the process by which water is carried through plants from roots and release water vapor through their leaves.

✿  Cutting down trees can decrease the amount of water vapor entering the air through the processes of <u>Transpiration</u>

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what is the force per meter on a straight wire carrying 5.0 a when it is placed in a magnetic field of 0.020 t
Dvinal [7]

The Force per meter on a straight wire carrying current in a magnetic field is<u>  0.045 N/m.</u>

<u>Calculation:-</u>

       F/ℓ = B I sin θ

  Where B – Magnetic field = 0.02 T I – Current = 5 A          

Substituting the values

F/ℓ = (0.02) (5) (sin 27 deg)

F/ℓ = <u>0.045 N/m</u>

A force is an influence that can alternate the motion of an item. A force can cause an item with mass to trade its pace, i.e., to boost up. force can also be described intuitively as a push or a pull. A pressure has both value and course, making it a vector quantity.

The push or pull on an item with mass causes it to change its velocity. force is an external agent capable of converting a frame's nation of relaxation or motion. It has significance and a path. A force is a push or pulls among gadgets. it is called an interplay because if one object acts on some other, its movement is matched with the aid of a reaction from the alternative object.

Learn more about force here:-brainly.com/question/12970081

#SPJ4

3 0
1 year ago
Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving
saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh

v_2^2=v_1^2-2gh

v_2^2=(7.7)^2-2\times 10\times 1

v_2=6.26\ m/s

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 *
marta [7]
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s

To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s

Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m

This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.

But anyways, good luck!



4 0
3 years ago
The diagram below shows a 5.00-kilogram block
bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

  • <em>Mass of the block, m = 5 kg</em>

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

  • <em>g is acceleration due to gravity = 10 m/s²</em>

W = 5 x 10

W = 50 N <em>(</em><em>downwards</em><em>)</em>

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0
2 years ago
Read 2 more answers
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
Assoli18 [71]

Answer:

Hence the pressure is 3\times 10^5 Pa

Explanation:

Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is 3\times 10^5 Pa

3 0
3 years ago
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