Is Mercury a heavier element than tin

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

3 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

3 years ago

Which most describes this table?

kati45 [8]

Answer:

In the table there we always eat.

It just based on my own...

3 0

2 years ago

Show all work.

lys-0071 [83]

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

$F=mg$ (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance $r$ from the centre of the Earth is given by

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

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

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

When the astronaut is on the Earth's surface, $r=R$ (where R is the Earth's radius), so his weight is

$F=\frac{GMm}{R^2}=640 N$

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

$r'=3R+R=4R$