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tia_tia [17]
2 years ago
13

You might say that this experiment was an attempt to build a scale, and then calibrate it against a scale that we trust (the ele

ctronic scale). So as not to give away the answer to the preceding question, let's assume that we're talking about a different scale with the same design. Several trials are conducted, and the average and 95% confidence interval are calculated. The mass determined using this other balance scale does NOT agree with the mass measured on a digital scale. Is it possible that the balance scale and the electric scale are accurate, and the lack of agreement was just random?
Physics
2 answers:
hammer [34]2 years ago
8 0

Answer:

yes

Explanation:

Allushta [10]2 years ago
3 0

No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

<h3>What is a random error?</h3>

Random error is defined as the deviation of the total error from its mean value due to chance.

Random errors can result from the instrument not being precise or from mistakes by the researcher.

Random errors can be minimized by taking multiple readings and averaging the results.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.

Learn more about random errors at: brainly.com/question/22041172

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In a classroom, 4 out of every 6 students are female. If there are 27 students, how many of them are female?
icang [17]

Answer:

16 male students, 11 female students

7 0
2 years ago
A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg =  0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0
3 years ago
Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0
2 years ago
Read 2 more answers
1HP is written in an electric motor. what does it mean?​
romanna [79]

Answer:

1HP is 1 Horse Power,

Explanation:

3 0
3 years ago
Read 2 more answers
You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
Monica [59]

a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

from above statement we got

height = 78.4 m

since the ball is thrown, so its vertical velocity would be zero

u = 0

taking g = 9.8m/s^2

now, using the equation of motion

h = ut + gt^2/2

now putting all the values in it

we got ,

78.4 = 9.8 * t^2/ 2

by solving we got,

t = 4 sec

b) now, since along the horizontal , no force acting and accelaration is zero so

R = ut , R is RANGE

R = 5 * 4

range =  20 m

c)  vertical components of the stone’s velocity just before it hits the ground = v sin θ =

horizontal   components of the stone’s velocity  just before it hits the ground = v cos θ

To know more about velocity  visit :

brainly.com/question/18084516

#SPJ9

4 0
1 year ago
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