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3 years ago

When designing an experiment, why is it important to test only one variable?

Physics

2 answers:

m_a_m_a [10]3 years ago

6 0

That's the only way you can get consistent and accurate results.

zheka24 [161]3 years ago

6 0

I will try to explain more in depth. Lets say Jimmy frequently has headaches and needs medicine to treat it. Jimmy uses Headfix and Achefix both in one experiment. The next day, Jimmy feels great. His headaches are cured, but since Jimmy used 2 variables (the 2 medicines), he is left with three options

a. Headfix worked
b. Achefix worked
c. Both worked.

Now Jimmy has no idea what solved his problem. It could be 3 possible answers. However, if he used one variable at a time and had two experiments instead, lets see what happens.

Jimmy uses Headfix.
a. It worked
b. It didn't work

Headfix did not work.

Next experiment Jimmy uses Achefix
a. It worked
b. It didn't work

Achefix worked.

Now instead of having three possible solutions, you have 1 definite answer experiment. Therefore you cannot use more than one variable because there could be multiple outcomes and you do not know which variable affected the results differently/

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Describe a situation where you add heat to a substance or material but there is no change in temperature. What does this look li

Alisiya [41]

ONE CAN perform this by doing an ideal experiment
by creating an isothermal system
its like you supply heat to a body and that body is present at very low temperature the amount of heat you supply is equal to the amount of heat lost by that body due to difference in the temperature of the body and the surrounding. heating curve will be constant as there is no change in the internal energy of the system ..

6 0

3 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$