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3 years ago

Velocity is:

Physics

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

Explanation:

Solution:-

- The Quantity of theory of money states:

M * V = P * Y

Where,

M = Money supply

V = Velocity of money exchange

P = The price level

Y = Real GDP

- By re-arranging the formula and solving for "V" we have:

V = P*Y / M

- The expression on right hand side increases if exchange of dollars increases.

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Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0

3 years ago

A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

3 0

3 years ago

For the circuit in the previous part, what happens to the maximum current if the frequency is doubled and the inductance is halv

tamaranim1 [39]

Answer:

Following are the responses to these question:

Explanation:

Since the $max^{m}$ is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance $(X_L)$ for capacities