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faltersainse [42]
3 years ago
10

A light wave travels through water (n = 1.33) at an angle of 35°. What angle

Physics
2 answers:
Dovator [93]3 years ago
4 0

Answer:1

Explanation:

1

Bas_tet [7]3 years ago
4 0

Answer:

30.6°

Explanation:

A P E X

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Physics help please! Offering 75 pts!
Nataliya [291]

Answer:

A

Explanation:

i got it from 2021 edge

5 0
2 years ago
Read 2 more answers
A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w
inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

6 0
3 years ago
If the rate $56 per 7 hours is reduced to a unit rate, the result is dollars per hour.
34kurt
The answer is 8 because multiplying 7 and 8 is 56
4 0
3 years ago
Read 2 more answers
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
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