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Svetradugi [14.3K]
3 years ago
7

All of the following are possible sources of error in a scientific investigation except for

Physics
2 answers:
kolezko [41]3 years ago
7 0

Your answer would be C my friend

dusya [7]3 years ago
4 0
"c" reviewing previously published studies before the experiment
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What happen when a star dies?
Vera_Pavlovna [14]
It depends on the size of the star. If it's size was normal then it cools down into White dwarf, then a black dwarf. If a really huge star dies, then we can see a "Supernova" from that.

Hope this helps!!
7 0
3 years ago
Scientists take scientific measurements carefully in order to ensure that reliability and validity. What is the difference betwe
Gala2k [10]
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3 years ago
If the Earth rotated more slowly about its axis, your apparent weight would
disa [49]
 <span>If the Earth rotated more slowly about its axis, your apparent weight would 
A) increase. 
B) decrease. 
C) stay the same. 
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8 0
3 years ago
You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m
Fofino [41]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0
3 years ago
Which of the following has mechanical energy?
KATRIN_1 [288]
Mechanical energy can have mechanical systems. The only mechanical system in the list is the compressed spring. A car battery and a glowing incandescent lightbulb have electrical energy, a nucleus of atom has potential (internal) energy.

5 0
3 years ago
Read 2 more answers
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