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3 years ago

How do you find a controlled variable

Physics

1 answer:

adelina 88 [10]3 years ago

7 0

Hello there!

Essentially, a control variable is what is kept the same throughout the experiment, and it is not of primary concern in the experimental outcome. Any change in a control variable in an experiment would invalidate the correlation of dependent variables (DV) to the independent variable (IV), thus skewing the results.

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Consider the data table charting the speed of a toy car moving across the floor. The line graph representing this data would BES

MrMuchimi

Answer:

The answer is D) diagonal line with varying slope, from 3 to 5 on USATestprep

Explanation:

4 0

3 years ago

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

STatiana [176]

Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

$P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2$

$1\ N/m^2=\frac{1}{6894.757}\ lb/in^2$

$3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2$

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0

3 years ago

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h

icang [17]

Answer:

Part a)

$P = 5.72 kg m/s$

Part b)

$\Delta P = 2.93 kg m/s$

Part c)

$F = 44.4 N$

Part d)

$\Delta P = 5.02 kg m/s$

Part e)

$\Delta t = 0.113 s$

Part f)

$\Delta K = 0$

Explanation:

As we know that initial velocity of the ball is given as

$v = 11.8 cos29 \hat i + 11.8 sin29 \hat j$

$v_i = 10.3 \hat i + 5.72 \hat j$

Now final velocity of the system is given as

$v_f = 10.3\hat i - 5.72\hat j$

Part a)

now magnitude of initial momentum is given as

$P = mv$

$P = 0.256(11.8)$

$P = 5.72 kg m/s$

Part b)

Change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(5.72 + 5.72)$

$\Delta P = 2.93 kg m/s$

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{2.93}{0.066}$

$F = 44.4 N$

Part d)

Magnitude of change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(7.8 + 11.8)$

$\Delta P = 5.02 kg m/s$

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

$\frac{\Delta P}{\Delta t} = F$

$\frac{5.02}{\Delta t} = 44.4$

$\Delta t = 0.113 s$

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

$\Delta K = 0$

8 0

3 years ago

In an emergency, a driver brings a car to a full stop in 5.0s. The car is traveling at a rate of 38m/s when the breaking begins.

Lana71 [14]

(a) By definition of average acceleration,

a = ∆v / ∆t = (0 - 38 m/s) / (5.0 s)

a = -7.6 m/s²

(b) Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the change in position. So

0² - (38 m/s)² = 2 (-7.6 m/s²) ∆x

∆x = 95 m

6 0

4 years ago

A car increases its speed from 4.20 m/s to 8.60 m/s over 3.20 seconds. What is the car’s acceleration?

katrin2010 [14]

Answer:1.375metre per second square

Explanation: acceleration=(final velocity-initial velocity)÷time

acceleration=(8.6-4.2)÷3.2

Acceleration=4.4÷3.2

Acceleration=1.375 metre per second square

5 0

3 years ago