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NikAS [45]
3 years ago
15

Which tool would a scientist use to find the mass of an object?

Physics
1 answer:
Luda [366]3 years ago
4 0
The common method would be to use Balance, If you think about it, on other planets, the balance weights change by the same factor as the object you are measuring. Your mass measured with a balance would be the same on the moon as it is on Earth. Weight is a measuring of gravity's effect on something. Mass is another story, it's the amount of matter in an object. Move to another planet and its object's weight will change, however, its mass will remain the same.

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Box A has two moles of gas at a temperature of 400 K. Box B has one mole of gas at a temperature of 800 K. Which of the two gas
mina [271]
B I think if not sorry
8 0
3 years ago
A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.
Bond [772]

Answer:

ax = -3.29[m/s²]

ay = -1.9[m/s²]

Explanation:

We must remember that acceleration is a vector and therefore has magnitude and direction.

In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.

a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]

3 0
2 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
Read 2 more answers
Who’s going faster in the attached image?
ioda

Answer:

Art

Explanation:

Polly's line is linear, while arts line is going up with constant velocity. There for art is going faster.

8 0
3 years ago
Read 2 more answers
The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares
KengaRu [80]

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, \theta_1=47.7^{\circ}

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

n_1\ sin\theta_1=n_2\ sin(90)

n_1\ sin(47.7)=1

n_1=\dfrac{1}{sin(47.7)}

n_1=1.35

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

5 0
3 years ago
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