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krek1111 [17]
3 years ago
7

What are two things you do ahead of time to prepare for a lab

Physics
2 answers:
Oxana [17]3 years ago
6 0
Wear your proper gear and gather all nessesary materials with instructors permission.
Fofino [41]3 years ago
3 0
-- Review the part of the book and your notes that deal with the lab topic.

-- Read the lab exercise so you know what you'll be doing in lab today.

-- Gather up everything you'll need to have there ... not the stuff like
chemicals or equipment that they'll give you in lab, but the stuff that
you're expected to bring; like pencils, lab notebook, scratch paper,
your calculator, a ruler, things like that.

-- Go to the bathroom.  
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A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
In this diagram, medium A represents water and medium B represents air. The phenomenon in this diagram, moving from A to B, is c
Readme [11.4K]
Its C) refraction
hope its may help you
6 0
3 years ago
Read 2 more answers
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
Properties can be observed only when the substance in a sample of matter are changing into different substance
Zolol [24]

The complete statement is "chemical properties can be observed only when the substance in a sample of matter are changing into different substance".


It states a key concept in chemistry.


A chemical property is the ability of a substance, element or compound, to <em>transform</em> into other substances either <em>by decomposing or by combining</em> with one or more substances.


This transformation is defined as chemical reaction.


During chemical reactions some chemical bonds  are broken and others are formed leading to the formation of one or more different substances called products.


Some examples of chemical properties are: reactivity with oxygen, reactivity with water, acidity, basicity, oxidation, reduction. The only way to tell if a substance has certain chemical property is by letting it react and so observe the change of the original substance into one or more different substances.

8 0
3 years ago
To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
3 years ago
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