Bob walks 290 m south, then jogs 630 m southwest, then walks 290 m in a direction 26 ∘ east of north.

In the bohr model, what is the ratio of its kinetic energy to its potential energy?

stich3 [128]

The centrifugal force C = mv^2/r = kq^2/r^2 = P centripetal force. m is the electron mass, q are the proton and electron charges (opposites), and r is the Bohr radius.

Thus 1/2 mv^2/r = 1/2 kq^2/r^2 and KE = 1/2 mv^2 = 1/2 kq^2/r = 1/2 PE

Therefore KE/PE = 1/2, no matter what state the electron is in.

8 0

3 years ago

What is Latent heat and also give types.<br>

dmitriy555 [2]

Answer:

Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process. Two common forms of latent heat are latent heat of fusion (melting) and latent heat of vaporization (boiling).

Explanation:

8 0

2 years ago

Give me four reasons pluto is a cool planet / dwarf planet

AleksandrR [38]

Answer:

1. Just because it is small doesn't mean it needs to be excluded. If that were the case I would've been out of my friend group a while ago

2. Look at it it's fricking beautiful (see attachment)

3. It is just a great planet I don't think there needs to be any reason given I meannn you agree?

4. There's no other reason needed it's fricking gorgeous and amazing it needs no other reason. It needs a freaking

opening announcement to announce the arrival of the gorgeouness.

5 0

3 years ago

Read 2 more answers

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

7 0

3 years ago

A ball is projected upward at time t = 0.0 s, from a point on a roof 60 m above the ground. The ball rises, then falls until it

musickatia [10]

<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here

then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))

The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>

7 0

3 years ago