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3 years ago

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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9 The diagram shows a uniform beam PQ. The length of the beam is 3.0 m and its weight is 50 N. The beam is supported on a pivot

tankabanditka [31]

equilibrium = 1/1

50 N/x = 1/1

x = 1/1 × 50 N

x = 50 N (B)

#LearnWithEXO

6 0

3 years ago

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

7 0

3 years ago

"Two electrons in an atom are separated by 1.4 × 10−10 m, the typical size of an atom. What is the force between them? The Coulo

Rudiy27

Answer:

$F\approx 1.17551\times 10^{-8} N$

Explanation:

In order to find the force between the two electrons, we need to use Coulomb's law. In its scalar form, the law is given by:

$F=k_c \frac{q_1*q_2}{r^2}$

Where:

$k_c=Coulomb\hspace{3}constant\approx 9\times10^{9} \frac{Nm^2}{C^2} \\q_1\hspace{3}and\hspace{3}q_2=Magnitudes\hspace{3}of\hspace{3}the\hspace{3}charges\\r=Distance\hspace{3}between\hspace{3}the\hspace{3}charges$

The electric charge of an electron is a known constant given by:

$q_e \approx -1.6 \times 10^{-19} C$

So:

$q_1=q_2=q_e \approx -1.6 \times 10^{-19} C$

Therefore, replacing the data provided in the Coulomb's law equation:

$F=(9\times 10^{9})\frac{(-1.6\times 10^{-19})*(-1.6\times 10^{-19})}{(1.4\times 10^{-10})^2} =1.175510204\times10^{-8} \approx 1.17551\times10^{-8}N$

8 0

3 years ago

A force of 15 newtons is applied to both Object A with a mass of 25 kilograms and Object B with a mass of 50 kilograms. What is

love history [14]

acceleration of object ais help the acceleration of an object b OD

3 0

3 years ago

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration du

natulia [17]

Given: Normal pull of gravity g = 9.8 m/s²;

g = 0.855 m/s² (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the Earth Me = 5.98 x 10²⁴ Kg

Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or

r = 21,610 Km

.

5 0

3 years ago