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Anna [14]
3 years ago
8

What type of tool seems to be the mist accurate for measuring liquids?

Physics
2 answers:
aliina [53]3 years ago
8 0

mesuring cup , beaker, graduated cylinder, burets, pipette or micropipette. <3





Anit [1.1K]3 years ago
6 0

Volumetric flasks are most accurate

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What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2
Margarita [4]

Answer:

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

Explanation:

  • We have to make a hollow sphere of inner  radius r_1 and outer radius r_2.

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

V_t=\frac{4}{3} \pi.r_2^3

And the volume of the hollow space in the sphere:

V_h=\frac{4}{3} \pi.r_1^3

Therefore the net volume of material required to make the sphere:

V=V_t-V_h

V=\frac{4}{3} \pi(r_2^3-r_1^3)

  • Now let the density of the of the material be \rho.

<u>Then the mass of the material used is:</u>

m=\rho.V

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

4 0
3 years ago
R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

3 0
2 years ago
A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t
Leona [35]

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

\int\limits {F} \, dx

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

\frac{dV}{dt} = a = 2.6

So we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, we replace everything:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

P = \frac{W}{t}

So

P = 161.9638/4.7 = 34.46 W

8 0
2 years ago
An object weighs 30 N on Earth. A second object weighs 30 N on the Moon. Which has the greater mass?
andreev551 [17]

Answer:

The object on earth, because earth has a bigger amount of gravity than how much the moon has.

Explanation:

6 0
2 years ago
Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a
I am Lyosha [343]

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

q_1\ and\ q_2 are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

F'=\dfrac{kq^2}{r'^2}

F'=\dfrac{kq^2}{(2r)^2}

F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}

F'=\dfrac{1}{4}\times 7.35

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

3 0
3 years ago
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