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3 years ago

15

Calculate the terminal velocity of

1 answer:

adell [148]3 years ago

6 0

I’m pretty sure it’s a

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Answer the question with step.

Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0

3 years ago

Nervous tissue makes up most of the

LiRa [457]

Answer:

, Nervous tissue is composed of two types of cells, neurons and glial cells. Neurons are the primary type of cell that most anyone associates with the nervous system. They are responsible for the computation and communication that the nervous system provides.

8 0

3 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.

LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

The moment of inertia of the system is 289.088 kg.m^2

7 0

3 years ago

How can we relate density and pressure in liquids ??

Masteriza [31]

Pressure increases with increasing depth. h2=2hh

6 0

3 years ago

Read 2 more answers

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

or

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago