Which of the following sentences uses so correctly?

In order to prevent injury in a car crash, it is recommended that you _______. Increase the initial velocity of the collision.

Vlad1618 [11]

Increase the change in momentum of the collision

5 0

3 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

Can someone please help me out with this quiz will give brainiest and thanks to people

Virty [35]

Answer:

Energy transferred = 28.8 Joules.

1. Energy transferred = 144 Joules.

2. The unit of potential difference, volts can also be described as Joules per Coulombs.

3. Current, I = 6.945 Amperes.

Explanation:

Given the following data;

Current, I = 1.2A

Time, t = 2 minutes

Potential difference, V = 12 volts.

To find the energy transfered;

Energy transferred = charge moved * potential difference

E = Q * V

Substituting into the equation, we have;

Energy transferred = (1.2 * 2) * 12

Energy transferred = 2.4 * 12

Energy transferred = 28.8 Joules.

1. <em><u>Given the following data;</u></em>

Charge, Q = 24C

Potential difference = 6V

To find the energy transferred;

E = Q * V

Substituting into the equation, we have;

E = 24 * 6

E = 144 Joules.

2. Since we know that, Energy transferred = charge moved * potential difference

$Potential \; difference = \frac {Energy \; transferred}{Charged \; moved}$

The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.

Therefore, the unit of potential difference becomes Joules per Coulomb.

3. <em><u>Given the following data;</u></em>

Potential difference = 18V

Energy transferred = 500J

Time, t = 4 minutes.

To find the current;

E = Q * V

Substituting into the equation, we have;

500 = Q*18

Q = 500/18

Q = 27.78C

But, Charge moved (Q) = current (I) * time (t)

Current, I = Q/t

Substituting into the equation, we have;

Current, I = 27.78/4

Current, I = 6.945 Amperes..

3 0

3 years ago

It is 5.00 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0 k

8_murik_8 [283]

Answer:

a. Walking burns up more energy.

b. 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

Explanation:

a. We know energy W = Pt where P = power and t = time.

Now for walking, t = d/v where d = distance = 5.00 km and v = speed = 3.00 km/hr and P = 290 W

So, t = d/v = 5.00 km/3.00 km/hr = 5/3 hr = 5/3 × 3600 s = 6000 s

W = Pt = 290 W × 6000 s = 1740000 = 1740 kJ

Now for running, t = d/v where d = distance = 5.00 km and v = speed = 10.00 km/hr

So, t = d/v = 5.00 km/10.00 km/hr = 0.5 hr = 0.5 × 3600 s = 1800 s and P = 700 W

W = Pt = 700 W × 1800 s = 1260000 = 1260 kJ

Since walking burns up 1740 kJ and running burns up 1260 kJ, walking burns up more energy.

b. It burns up 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

4 0

4 years ago

ANSWERS UNDER QUESTIONS:

Julli [10]

<h2>Answers:</h2><h2 /><h2>a) Arrow B</h2><h2>b) Arrow E</h2>

Explanation:

Refraction is a phenomenon in which a wave (the light in this case) bends or changes its direction <u>when passing through a medium with a refractive index different from the other medium.</u> Where the Refractive index is a number that describes how fast light propagates through a medium or material.

According to this, if we observe the rays A an D passing throgh the biconcave lens, we will have two mediums:

1) The air

2)The material of the biconcave lens

This two mediums have different refractive indexes, hence the rays will change the direction.

-For the incident ray A, the corresponding refractive ray is B, because is the ray that bends after passing throgh the lens

-For the incident ray D, the refracted ray is E following the same principle.

6 0

3 years ago

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