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3 years ago

What apparatus can be used to measure the internal diameter of a test tube?

Physics

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

To determine the volume of a given beaker/calorimeter by measuring its internal diameter and depth with vernier calipers. A vernier caliper is a measuring instrument with two scales: a main scale and a vernier scale that slides over the main scale.

Explanation:

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A 2-kg ball is moving with a speed of 4 m/s, and a 4-kg ball is moving with a speed of 2 m/s. What can you conclude about the of

Mila [183]

<h2>Kinetic energy of mass 4 kg ball is less than kinetic energy of mass 2 kg ball</h2>

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

For ball of mass 2 kg

Mass, m = 2 kg

Velocity, v = 4 m/s

Kinetic energy = 0.5 x 2 x 4² = 16 J

For ball of mass 4 kg

Mass, m = 4 kg

Velocity, v = 2 m/s

Kinetic energy = 0.5 x 4 x 2² = 8 J

Kinetic energy of mass 4 kg ball is less than kinetic energy of mass 2 kg ball

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3 years ago

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

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Answer:

$H_2 = 91.55 km$

Explanation:

Gravity on the surface of planet is given as

$g = \frac{GM}{R^2}$

as we know that

$M = 8.93 \times 10^{22} kg$

$R = 1821 km$

now gravity on the planet is

$g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}$

so we have

$g = 1.8 m/s^2$

now we know that

$H_{max} = \frac{v^2}{2g}$

so we will say

$\frac{H_1}{H_2} = \frac{g_2}{g_1}$

$\frac{500}{H_2} = \frac{9.81}{1.8}$

$H_2 = 91.55 km$

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3 years ago

The mysterious visitor that appears in the enchanting story The Little Prince was said to come from a planet that "was scarcely

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Answer:

Part a)

$g = 1 \times 10^{-5} m/s^2$

Part b)

$v = 0.011 m/s$

Explanation:

Part a)

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

$g = \frac{G(\frac{4}{3}\pi R^3 \rho)}{R^2}$

$g = \frac{4}{3}\pi \rho G R$

now we know that

$\rho = \frac{M}{\frac{4}{3}\pi R^3}$

$\rho = \frac{5.98 \times 10^{24}}{\frac{4}{3}\pi (6.37 \times 10^6)^3}$

$\rho = 5523.2 kg/m^3$

now we have

$g = \frac{4}{3}\pi (5523.2) (6.67 \times 10^{-11})(6.5)$

$g = 1 \times 10^{-5} m/s^2$

Part b)

As we know that the escape speed is given as

$v = \sqrt{\frac{2GM}{R}}$

$v = \sqrt{2gR}$

$v = \sqrt{2(1\times 10^{-5})(6.5)}$

$v = 0.011 m/s$

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Answer:

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Explanation:

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What apparatus can be used to measure the internal diameter of a test tube?​

What apparatus can be used to measure the internal diameter of a test tube?