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Morgarella [4.7K]
1 year ago
10

A scientist decides to replicate an experiment completed by another scientist. Which statement describes something that would no

t affect the results of the replicated experiment?
A) The two experiments were completed by two different people.
B) The variables were not kept the same in both experiments.
C) The two scientists did not use the same equipment.
D) There were some measurements that were different in the two experiments.
Physics
1 answer:
tatyana61 [14]1 year ago
6 0

The statement describes something that would not affect the results of the replicated experiment is option A. (The two experiments were completed by two different people)

<h3>What is an experiment?</h3>

An experiment can be defined as a procedure carried out to support or refute a hypothesis, or determine the efficacy or likelihood of something previously untried.

Experiments helps in the provision of insight into the cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.

If two experiments were completed by two different people, it would not affect the results if the experiment is replicated.

The purpose of any experiment is to test or verify a hypothesis

Learn more about experiments at: brainly.com/question/26117248

#SPJ1

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when an electric current is flowing through a wire which of the following is also produced energy domain field force
qaws [65]

When an electric current is flowing through a wire, there is a magnetic <em>field</em> produced.  The lines of magnetic force are circles around the wire.

4 0
4 years ago
A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick, at right angles to the stick, with a
katen-ka-za [31]

Answer

given,

mass of the stick = 290 grams = 0.29 Kg

Force on the stick on one side = F =  9 N

force acting perpendicular to stick.

magnitude of acceleration

rate of change of angular momentum is equal to Force

rate of change of angular momentum = 9 N

F = m a

a = \dfrac{F}{m}

a = \dfrac{9}{0.29}

a = 31.034 m/s²

Direction of motion will in the direction of force application or in the direction of change of velocity

5 0
4 years ago
The speed for a car that went a distance of 125 miles in 2 hours time is ______. Question 2 options: 250 meters/sec 62.5 miles/h
Goryan [66]

Answer:

62.5 miles per hour

Explanation:

Speed = Distance travelled / Time taken

Speed = 125/2 = 62.5

You derive the units of the speed...by using the speed formula....,

Speed = Distance/Time

Speed = miles/hour

Hence the units for the speed = miles/hour

5 0
3 years ago
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
3 years ago
Bone has a Young's modulus of about
saveliy_v [14]

Answer: 4.74 mm

Explanation:

We can solve this problem with the following equation:

Y=\frac{stress}{strain} (1)

Where:

Y=1.8(10)^{10} Pa is the Young modulus for femur

stress=\frac{F}{A}=1.58(10)^{8} Pa is the stress (force F applied per unit of transversal area A) on the femur

strain=\frac{\Delta l}{l_{o}}

Being:

\Delta l the compression the femur can withstand before breaking

l_{o}=0.54 m is the length of the femur without compression

Writing the data in equation (1):

Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l_{o}}} (2)

1.8(10)^{10} Pa=\frac{1.58(10)^{8} Pa}{\frac{\Delta l}{0.54 m}} (3)

Isolating \Delta l:

\Delta l=\frac{(1.58(10)^{8} Pa)(0.54 m)}{1.8(10)^{10} Pa} (4)

\Delta l=0.00474 m (5) This is the compression in meters

Converting this result to millimeters:

\Delta l=0.00474 m \frac{1000 mm}{1 m}=4.74 mm

4 0
4 years ago
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