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pychu [463]
3 years ago
12

What happens to the magnitude of the force of gravitation between two objects if: 1. distance between the objects is tripled? 2.

mass of both objects doubled? 3. mass of both objects as well as the distance distance between them is doubled? EES CLASS TEACHER GRADE 9 A Oct 5 3 class comments
Physics
2 answers:
Ludmilka [50]3 years ago
8 0
<span>2. mass of both objects doubled? 
Hope it helped!</span>
VashaNatasha [74]3 years ago
3 0
2. mass of both objects doubled, because with two negatives makes a positive in resulting of my answer....(this formula i made up, i use it on problems like this, i feel so accomplished)
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If a voltmeter has a less than ideal resistance, say 1 MΩ, and is used to measure the voltage across a resistor of a comparable
Naddik [55]

Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

lets suppose the supplied voltage is 1 V.

First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

R = 0.33 MΩ

As you can see the voltmeter will see a reduced resistance and the corresponding voltage also reduces because resistance and voltage are directly proportional.

Therefore, it is preferred to have a very high parallel resistance of the voltmeter.

8 0
2 years ago
A woman lifts her 100-newton child up one meter and carries her for a distance of 50 meters to the child's bedroom. How much wor
tankabanditka [31]
100 J

Please mark me brainliest it would be greatly appreciated haha
5 0
3 years ago
Read 2 more answers
Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u
tino4ka555 [31]

Answer:

a)  force between them is attraction,   b)  F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

          F =k \frac{q_1q_2}{r^2}

           

In the exercise they give us the values ​​of the loads

          q1 = - 10 mC = -10 10⁻³ C

          q2 = 5 mC = 5 10⁻³ C

           d = 20 cm = 0.20 m

let's calculate

          F = 9 10⁹ \frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}

          F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0
3 years ago
Gabriella got a burn on her arm from a chemical spill and rinsed the area thoroughly with water. Which piece of safety equipment
ioda
I believe it she should use the first aid kit next
8 0
3 years ago
Read 2 more answers
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

Change in entropy \Delta s = m(s_2 - s_1)

              =3( 3.36035 - 2.88) =  1.44 kJ/kg

8 0
3 years ago
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