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4 years ago

Is there an atom within an atom?

1 answer:

6 0

No. An atom is made up of 3 kinds of particles, protons, neutrons, and electrons. Protons and neutrons make up the center of the atom.

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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

g Which of the following wavefunctions are: a) square-integrable on the interval provided (1 point); b) valid wavefunction satis

Goshia [24]

Answer:

The answer is given in the attachment

Explanation:

5 0

4 years ago

A 12.0-g cd with a radius of 5.81 cm rotates with an angular speed of 33.2 rad/s. what is its kinetic energy?

schepotkina [342]

Use the formula 1/2*I*w^2 where w is the angular speed and I is the rotaional inirtia (for disk it is .5*mass*radius^2).

4 0

3 years ago

A water skier is pulled behind a boat by a rope. The rope has a tension of 650 N and is at an angle of 27°. What is the y-compon

Irina-Kira [14]

There is no figure, so I asume the angle of the rope is measured with respect to the horizontal direction.

If this is the case, then the magnitude of the tension is the hypothenuse of a right triangle, of which $T_x$ and $T_y$ (horizontal and vertical component of the tension) are the other sides, and $\alpha$ is the angle between T and $T_x$ .
Therefore, $T_y$ (the side of the triangle opposite to $\alpha$ ) is given by the magnitude of the tension multiplied by the sine of the angle:
$T_y = T \sin \alpha = (650 N)(\sin 27^{\circ} )=295 N$

4 0

3 years ago

Read 2 more answers

The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc

Alona [7]

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

Explanation:

First we will find the acceleration of pickup truck.

As, the acceleration is uniform, therefore we can use Newton's third equation of motion:

2as = $V_{f}^{2}-V_{i}^{2}$

First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

therefore,

37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr = 54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

F = (460 lbm +72 lbm)(6.88 ft/sec²)

F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

Force in each peg = F/2 = (3660 lbf)/2

<u>Force in each peg = 1830 lbf</u>

4 0

3 years ago