The concentration of water vapor in the atmosphere known as

An atom has 25 protons 30 neutrons and 25 electrons what is the charge of the atom nucleus

Alex17521 [72]

The nucleus is always positive since thats where the protons are located and the neutrons have no charge....

3 0

3 years ago

A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1

Ilia_Sergeevich [38]

Answer:

1.5 m/s

Explanation:

Conservation of momentum means the momentum of the system before the collision is the same as after.

The before, after momentum of each ball is ...

5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)

10 kg ball: (10 kg)(0 m/s), (10 kg)(v)

The sum of the "before" products is the same as the sum of the "after" products:

(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v

(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides

v = (15 kg·m/s)/(10 kg) = 1.5 m/s

The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.

7 0

3 years ago

What is the wavelength of radiation with a frequency of 1.50x10^13 hz?

stiks02 [169]

Π = c/ f
π = 3×10^8 / 1.5×10^13
π = 3÷1.5 ×10^(8-13)
π = 2×10^-5 m

7 0

4 years ago

An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce

Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

$V_i = \pi R^2 h_i --- (1)$

where;

height h which is the height for the free surface in a rotating tank is expressed as:

$h = \dfrac{\omega^2 r^2}{2g} + C$

at the bottom surface of the tank;

r = 0, h = 0

∴

$h = \dfrac{\omega^2 r^2}{2g} + C$

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

$h=\dfrac{\omega^2 r^2}{2g} --- (2)$

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

$\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr$

replacing the value of h in equation (2); we have:

$V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)$

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then $V_f = V_i$

Replacing equation (1) and (3)

$\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i$

$\omega^2 = \dfrac{4g \times h_i }{R^2}$

$\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}$

$\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }$

$\omega = \sqrt{109.87 }$

$\mathbf{\omega = 10.48 \ rad/s}$

Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s

6 0

3 years ago

Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed

quester [9]

Answer:

F = - 1,598 10⁻³ N

Explanation:

Electic strength is given by Coulomb's law

F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

q₀ = 4.0 10⁻⁶ C

Let's calculate

F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

F = - 1,598 10⁻³ N

4 0

4 years ago