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3 years ago

6

Helppp plzzzz.......A motor home starts across a lake. The lake is 7 miles long and it takes 0.5 hours to get across. What is th

e speed? Write a formulae and show a label.

1 answer:

aleksklad [387]3 years ago

7 0

Speed=distance\time so try 14??

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A stone is dropped off a cliff and it falls for 4.8 m/s

valina [46]

Answer:

you cant caculate this because there is no question

Explanation:

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3 years ago

How long does it take to fall from 5000 feet?

WINSTONCH [101]

It depends on your weight, your hieght, and how fast you are falling

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4 years ago

The bending of a wave as it enters a new medium at an angle is called _____.

Andru [333]

Answer:

C). Refraction

Explanation:

When light enters from one medium to another medium then it changes it's direction from its initial direction

the phenomenon of changing the direction due to change in the medium is known as Refraction of light

So here we know that when light strikes an interface which separates two medium then due to change in the medium we will have

$\mu_1 sin\theta_1 = \mu_2 sin\theta_2$

so here we have

$\theta_1, \theta_2$ = two angles of the light in two mediums

$\mu_1 , \mu_2$ = two refractive index of the two mediums

So correct answer will be

C). Refraction

4 0

3 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (<em>the train must see the car advancing at a lower speed</em>), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (<em>relative to the ground</em>):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, <u>we have to do all the same</u> but using $v_{AT}=v_A+v_T$ (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

4. How far apart are a proton and electron if they exert an attractive force of 3 N on one another?

alexandr402 [8]

Answer:

$8.76\cdot 10^{-15}m$

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

And the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

F = 3 N is the force between the proton and the electron

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and the electron

Therefore, solving for r, we find the separation between the two particles:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{3}}=8.76\cdot 10^{-15}m$

5 0

3 years ago