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Elina [12.6K]
3 years ago
5

Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T

hinking only about the first part of the question, mass, what might be a hypothesis that would illustrate the relationship between mass and kinetic energy? Use the format of "if...then... because when writing your hypothesis
Physics
2 answers:
Natalka [10]3 years ago
9 0

Answer:

If the mass of an object increases, then its kinetic energy will increase proportionally because mass and kinetic energy have a linear relationship when graphed.

Explanation:

Taya2010 [7]3 years ago
7 0

Explanation:

The kinetic energy is said to be possessed due to the motion of the object. An object at rest will have zero kinetic energy and if it is in motion it will have some kinetic energy. The mathematical expression for kinetic energy is given by :

E=\dfrac{1}{2}\times m\times v^2...........(1)

Where

m is the mass of the object

v is the velocity of object

It is clear form expression (1) that the kinetic energy of the object is directly proportional to the mass and velocity of an object.

So, the hypothesis for the mass and kinetic energy can be written as " when the mass of the object increases, its kinetic energy also increases because there exists a direct relationship between the mass and the kinetic energy of the object".

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Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0
2 years ago
A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove
jonny [76]

We must remember that the total net force equation at constant velocity is:

<span>F – Ff = 0</span>

of

F - µN = 0

Using Newton's 2nd Law of Motion:<span>

F = m a 

<span>Where,

F = net force acting on the body 
m = mass of the body 
a = acceleration of the body 

Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to 

F = net Force = 0 

Therefore, 

F - µN = 0 

where 

µ = coefficient of friction = 0.20 
N = normal force acting on the cart = 12 N 

Therefore, 

F - 0.20(12) = 0 

<span>F = 2.4 N </span></span></span>
4 0
3 years ago
A parallel circuit contains an 18-V battery wired with 2 bulbs with resistances of 8
RSB [31]

Answer:

See below

Explanation:

Total current will be   18 v/ 8 ohms +   18v / 24 ohms = 3 amps

Equivalent resistance   =   1 / (1/8 + 1/24) = 6 Ω

7 0
2 years ago
A coil with 55 loops has its flux change from 0 Wb to 0.0266 Wb in a certain amount of time, generating 5.22 V of EMF. How much
Eduardwww [97]

Answer:

0.280 s

Explanation:

I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.

4 0
3 years ago
An object moves in uniform circular motion what is true regarding the force on the object
PtichkaEL [24]
The force on the object has a constant strength, but its direction
keeps changing.  The force is always directed from the object to
the center of the circle.  It's called "centripetal force".
4 0
4 years ago
Read 2 more answers
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