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3 years ago

Is the mass of an object measured with scale, yes or no

Physics

1 answer:

mamaluj [8]3 years ago

7 0

Answer:

No.

Explanation:

The scale actually measures the force that the object does against it, and that force is called the weight.

Such that if we have an object with mass M and we are on Earth, where the gravitational acceleration is g, the weight is:

W = M*g

Now, there is a unit called "kilogram-force"

Such that on Earth, an object that has a mass of 10 kilograms, weighs 10 kilograms-force.

Then from the weight measure, we can instantly know the mass of the object, but the thing that is being measured is the weight, not the mass.

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th

Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0

3 years ago

Scientific models can be used for a variety of different purposes. Which of the following is not a possible use for a scientific

alex41 [277]

Options a to c can be the reasons for scientific models.

But to primarily answer scientific questions,that would require an empirical and experimental approach and not use of models.

Though after getting the answers, models can be built to further explain the answers.

<span>d. answer scientific questions.</span>

4 0

3 years ago

Read 2 more answers

Which object has the least amount of Kinect energy

Angelina_Jolie [31]

Answer: Could you please add the answer choices.

Explanation:

Thank you :)

6 0

3 years ago

Is the mass of an object measured with scale, yes or no​

Is the mass of an object measured with scale, yes or no