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ycow [4]
3 years ago
10

Humidity can be measured using human _____.

Physics
2 answers:
Mkey [24]3 years ago
8 0
Can be measured by using humans hair
Nutka1998 [239]3 years ago
6 0
It says on a website that humidity can be measured by human hair.
You might be interested in
If two objects have the same mass, what determines the strength of the gravitational force between them? (4 points) Select one:
Evgen [1.6K]

According to this equation

F = G × m₁*m₂ ÷ r²

other than the mass, the distance also affects the gravitational force between two objects (same mass or not).

Therefore the correct answer is B. The distance between the objects

Future note* use formulas to help you figure these sort of questions out. (if they have a formula to begin with).

3 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
dimaraw [331]

Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

7 0
3 years ago
Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
marissa [1.9K]

Answer:

maximum static frictional force of the wall on the book (Increasing)

normal force of the wall on the book (Decreasing)

weight of the book (Not changing)

Explanation:

Now according to Newton's third law of motion

"Every action has equal but opposite reaction"

By the data given in question, Albert was pressing the book against the wall.Now, Albert started to reduce his force up against the wall.

First we have to consider all the forces applied on book in this scenario.

1. Weight of book acting downwards (y-axis)

2. Friction between book and wall acting upward (y-axis)

3. Albert's force on book against wall (x-axis)

4. Normal reaction of wall against Albert's force (x-axis)

Now, when Albert reduced his force, new scenario will be

1. Weight will be remain constant as it is W = mg

Neither mass nor acceleration due to gravity changed, so weight acting upon the book will remain same.

2. When Albert reduced force, normal reaction of wall reduced against it according to Newton's third law of motion

3. Now notice that friction is a force which acts in accordance with the applied force. For example if a box is placed at floor, no friction is applied, but when you drag the box, friction starts to act and increases until its limit comes. So, when Albert reduced his force, weight will try to pull the book and maximum static friction will increase to hinder the movement of book downwards.

Notice that maximum static friction will hinder the book from movement, since Albert reduced his force, but wight didn't pull the book, which means that maximum static friction increased to hinder downward motion.

7 0
3 years ago
If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?
aksik [14]

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

8 0
3 years ago
You are sitting on a deck of your house surrounded by oak trees. You hear the sound of an acorn hitting the deck. You wonder if
Black_prince [1.1K]

Answer: 96N

Explanation:

To calculate the velocity of the impact On the persons head, we have

h = gt²/2

14 = 9.81t²/2

t² = 28/9.8

t² = 2.86

t = 1.69s

V = u + at

V = 0 + 9.81*1.69

V = 16.58m/s

a(average) = (v1² + v2²) /2Δy

a(average) = 16.58² + 0)/2 * 0.005

a(average) = 274.8964/0.01

a(average) = 27489.64m/s²

Using newton's second law of motion,

F(average) = m * a(average)

F(average) = 0.0035 * 27489.64

F(average) = 96.21N

Therefore the force needed by the acorn to do much damage starts from 96N

8 0
3 years ago
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