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3 years ago

Dakdadakdadakdadakda

Physics

1 answer:

notsponge [240]3 years ago

4 0

Answer: dakdadakdadakdadakda

Explanation:(sings) blah blah blah middle fingers in the air l-l-l-loser

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A 4.8 mF capacitor in series with a 500 Ω resistor is connected, by a switch, to a 12 V battery. The current through the resisto

ad-work [718]

Answer:

Current will be 81.7 mA

Which is not given in bellow option

Explanation:

We have given capacitance $C=4.8mF=4.8\times 10^{-3}F$

Resistance R = 500 ohm

Voltage V = 12 volt

We know that time constant of RC circuit of RC circuit is given by

$\tau =RC=500\times 4.8\times 10^{-3}=2.4sec$

Time is given as t = 1 sec

We know that current in RC circuit is given by

$i=\frac{v}{R}(1-e^{\frac{-t}{\tau }})$

So current $i=\frac{12}{500}(1-e^{\frac{-1}{2.4 }})=0.00817A=81.7mA$

Which is not given in the following option

3 0

3 years ago

Read 2 more answers

A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist

marshall27 [118]

Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= $\frac{71\times10^3}{4 \times \pi\times(220)^2}$

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.

3 0

3 years ago

Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium

ivolga24 [154]

Answer:

Explanation:

Iron and gadlinium are both very easily made into magnetic substances. Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

5 0

3 years ago

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$