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pentagon [3]
3 years ago
13

Davien and Jackson are headed to a beach that is 50km from school. At noon, Davien leaves his house that is 10 km closer to the

beach than the school is and moves at 40 km/h. Jackson starts from school at 12:30 P.M. And moves at 100 km/h. What time does Davien reach the beach? * Answer Format: Time
Physics
1 answer:
snow_tiger [21]3 years ago
5 0

Distance between Davien's school and beach, D = 50 km.

Also, Davian's house is 10 km closer to the beach than the school.

So, distance between Davien's house and beach, d = 40 km.

Speed of Davian, s = 40 km/h.

Now, time taken is given by :

t =\dfrac{Distance}{Speed}\\\\t=\dfrac{40}{40}=1\ hour

Therefore, Davien reach the beach at 1 : 30 P.M .

Hence, this is the required solution.

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nlexa [21]

Answer:

A spectroscope

Explanation:

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6 0
3 years ago
Which effect is created when two sound waves that are close in pitch
Temka [501]

Answer:

C. Beats

Explanation:

When waves are interfering with each other, the sound is louder in some places and softer in others. As a result, we hear pulses or beats in the sound.

5 0
4 years ago
Water flows over a section of Niagara Falls at the rate of 1.2×106 kg/s and falls 50.0 m. How much power is generated by the fal
quester [9]

Answer:

5.88×10⁸ W

Explanation:

Power = energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

7 0
3 years ago
which of this is not example of simple machine? A) a crowbar,B)a knife,C)a bicycle,D)a wheelchair ramp
DaniilM [7]
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3 0
3 years ago
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

Explanation:

Since R(t)=\frac{V}{I(t)}, we calculate the resistance rate by deriving this formula with respect to time:

\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})

Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega

8 0
3 years ago
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