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2 years ago

What is the volume (in cm3) of a 36.7 g piece of metal with a density of 6.95 g/cm3?

Physics

1 answer:

Basile [38]2 years ago

3 0

To calculate the volume of given metal, we use the formula

$Density= \frac{mass}{volume}$

Given, the mass of metal 36.7 g and density of metal is 6.95 g/cm3

Therefore,

$6.95 g/cm^{3} = \frac{36.7 g}{volume}$

or $volume = \frac{36.7 g}{6.95 g/cm^{3} }$

$volume = 5.3 cm^{3}$

Thus, the volume of metal piece is $5.3 cm^{3}$

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2 years ago

You are driving straight, travelling at 25 mph. You stomp on the brakes and come to a complete stop in a distance of 15 m. Assum

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Answer:

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2 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

Furkat [3]

Answer:

a) $F = 78.606\,N$ , b) $F = 88.911\,N$

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m \cdot g \cdot \sin \theta$

$F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}$

$F = 78.606\,N$

b) The equations of equilibrium are the following:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m\cdot (a + g \cdot \sin \theta)$

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$F = 88.911\,N$

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2 years ago

¿ Cuáles son los principales aportes de las ciencias en el desarrollo del mundo? Menciona una aplicación de Física del contenido

maria [59]

Answer:earth, water, air, fire, and (later) aether, which were proposed to explain the nature and complexity of all matter in terms of simpler substances.

Explanation:

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2 years ago

The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

Kipish [7]

Answer:

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Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

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$T=2\pi \sqrt{\frac{l}{g}}$ (1)

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T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

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Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

$T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}$

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

$g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}$

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

5 0

2 years ago