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2 years ago

9

Why does increasing the number trials increase confidence in the results of the experiment?

2 answers:

IceJOKER [234]2 years ago

6 0

If you increase the number of trials in an experiment it will make the test more valid and legitimate.As you take the same test/experiment once or twice you could see if your results are similar to each other.

juin [17]2 years ago

5 0

Increasing the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value. {This is the simple answer, That I used and got it correctly}

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If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t

Whitepunk [10]

From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer: 283 s (nearest second)

6 0

2 years ago

Acetone, a component of some types of fingernail polish, has a boiling point of 56°C. What is its boiling point in units of kelv

mixas84 [53]

Answer:

The boiling point of Acetone is 329K (in 3 significant figures)

Explanation:

Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)

7 0

2 years ago

Read 2 more answers

An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving

liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0

2 years ago

A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen

saw5 [17]

Answer:

The electric field strength is $4.5\times 10^{4} N/C$

Solution:

As per the question:

Area of the electrode, $A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}$

Charge, q = 50 nC = $50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m$

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

$\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}$

Now, the electric field strength of the electrode is:

$\vec{E} = \frac{\sigma}{2\epsilon_{o}}$

where

$\epsilon_{o} = 8.85\times 10^{- 12} F/m$

$\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}$

$\vec{E} = 4.5\times 10^{4} N/C$

7 0

2 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0

2 years ago