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3 years ago

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A ball is revolving horizontally in a circle of radius 1.8 m, and is held by a rigid, massless rod. The mass of the ball is 0.1

kg. The path of the ball has a constant linear speed of 27 m/s. What is the angular velocity of the orbit in rad/s?

1 answer:

OLga [1]3 years ago

4 0

Answer:

$\omega=15 \ rad/s$

Explanation:

given,

radius of circle = 1.8 m

mass of the ball = 0.1 Kg

linear speed of the ball = 27 m/s

angular velocity of orbit = ?

$v = r \omega$

ω is the angular speed of the circle

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{27}{1.8}$

$\omega=15 \ rad/s$

the angular velocity of the orbit is $\omega=15 \ rad/s$

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Y_Kistochka [10]

Answer:

B. Wave-like way with a pattern that is wave-like

Explanation:

The double slit experiment when performed with electromagnetic waves, gives a pattern of light lines and dark areas, equally spaced.

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3 years ago

If you dropped a rock and it fell 4.9 m in 1 s , how far would it fall in 3 s

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The formula relevant for this is:

h = v0t + 0.5 gt^2

since the rock was dropped, therefore:

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we can see that:

h / t^2 = 0.5 g = constant

therefore:

4.9 m / (1 s)^2 = h / (3 s)^2

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4 years ago

At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

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The rate of change of this distance is calculated as;

r² = a² + b²

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3 years ago

While taking a shower, you notice that the shower head is made up of 44 small round openings, each with a radius of 2.00 mm. You

Julli [10]

To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.

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$A = \pi r^2$

$A = pi (2*10^{-3})^2$

$A = 12.56*10^{-6}m^2$

At the same time the rate of flow would be

$Q = \frac{1L}{2s}$

$Q = 0.5L/s = 0.5*10^{-3}m^3/s$

By definition the discharge is expressed as

$Q = NAv$

Where,

A= Area

v = velocity

N = Number of exits

Q = NAv

Re-arrange to find v,

$v = \frac{Q}{NA}$

$v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}$

$v = 0.9047m/s$

PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2$

Replacing with our values we have that

$1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)$

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$v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}$

$v_1 = 3.54097m/s$

PART C) Assuming that water is an incomprehensible fluid we have to,

$Q_{pipe} = Q_{shower}$

$v_{pipe}A_{pipe}=v_{shower}A_{shower}$

$3.54097*A_{pipe}=0.9047*12.56*10^{-6}$

$A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}$

$A_{pipe = 3.209*10^{-6}m^2$

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3 years ago