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slamgirl [31]
4 years ago
8

What are the STEPS to using a triple beam balance

Physics
1 answer:
Gwar [14]4 years ago
8 0
A good diet. and exersise is the answer
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(a) Are the radiating electric field lines around a charged particle straight lines when the particle is stationary?
s2008m [1.1K]

Answer:

Yes

Explanation:

(a)

Yes, the radiating electric field lines around a stationary charged particle are straight lines when the particle is stationary.

7 0
3 years ago
You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of
Rzqust [24]

Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

                                       = 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0
3 years ago
9ma electric current is flowing through a conducting wire , then the number of electron passing through it in 3 min is
Leviafan [203]

Answer:

1.0125 x 10^19

Explanation:

current flowing through conductive wire= 9mA = 9 x 10^ -3 A

charge passing per 3 min

Q = It

= 9 x 10^ -3 x (3 x 60)

= 1.620 C

no of electrons in charge

Q = ne

1.620 = n x 1.6 x 10 ^ -19

n. = 1.0125 x 10 ^19

4 0
3 years ago
1. What are three questions you should ask before joining a club?
lara [203]

Answer:

what time does it start.

what do I need to join.

what are your expectations.

7 0
3 years ago
Consider a corn yield of 7,500 kg/hectare (equivalent to 120 bushels per acre). If 25 kg (one bushel) of corn consumed about 20m
Alla [95]

Answer:

\frac{W_{1}}{W_{2}} = 0.5

Explanation:

The ratio of the weight of corn to the weight of water consumed will be the same. Either we compare the total weights for the total amount of corn or we compare only a small proportion of it. So, we consider the given 25 kg of corn. So, the mass of corn is given:

m₁ = mass of corn = 25 kg

Now, for mass of water, we will use its density:

ρ₂ = m₂/V₂

m₂ = ρ₂V₂

where,

m₂ = mass of water = ?

V₂ = Volume of water consumed by 25 kg corn = 20 m³

ρ₂ = density of water = 1000 kg/m³

Therefore,

m₂ = (1000 kg/m³)/(20 m³)

m₂ = 50 kg

Now, we calculate the ration:

\frac{Weight of Corn}{Weight of Water} = \frac{W_{1}}{W_{2}} = \frac{m_{1}g}{m_{2}g}

\frac{W_{1}}{W_{2}}=\frac{m_{1}}{m_{2}} = \frac{25 kg}{50 kg}

\frac{W_{1}}{W_{2}} = 0.5

6 0
3 years ago
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