Assume that a cloud consists of tiny water droplets suspended (uniformly distributed,

10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.

oksano4ka [1.4K]

The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1

8 0

3 years ago

The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf

Galina-37 [17]

Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| 
E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J 

After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:

E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m 

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum

7 0

3 years ago

Which part of earths surface has the greatest rotational speed

Anni [7]

Every point on the surface must have the same rotational speed.
Otherwise some places would rotate away from other places.

If the next block of your city rotated faster than the block that you live on,
then you could sit at home, look out the window, and watch your school
rotate past your house.

The map of the continents on the Earth would change constantly.

6 0

3 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago

Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump.

earnstyle [38]

Answer:

2,87 * $10^{-3}$

Explanation:

When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:

Px= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

Py= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).

For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.

Pz = 1,5953 = m * 554

m = 2,87 * $10^{-3}$ kg

8 0

3 years ago