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KengaRu [80]
4 years ago
5

Derek runs 4 laps around the track. If each lap around the track is 0.25 miles long, and he starts and stops in the same locatio

n, what is Derek's displacement?
Physics
2 answers:
trapecia [35]4 years ago
6 0
Derek's displacement is 0 miles, because he starts and stops in the same place. The distance he covered would be 1 mile.

Distance is the total length covered.
Displacement is the distance between the starting point and the end point. If Derek started and stopped in the same place, he had a displacement of 0.

Hope this helps!
Sergeeva-Olga [200]4 years ago
5 0
1.00 1 mile      1 mile 1.00
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Your GPS shows that your friend’s house is 10.0 km away. But there is a big hill between your houses and you don’t want to bike
kati45 [8]

Answer:

The value is c  = 8 \  km

Explanation:

From the question we are told that

The distance of friends house from your point is a =  10 \  km

The distance of your friends street from your street is b =  6 \  km \  in the \ direction \  towards \  the  \  north

The diagram illustrating this question is shown on the first uploaded image

From the diagram we can apply by Pythagoras theorem as follows

a^2 =  b^2 +  c^2

=>     c  =  \sqrt{^2 -  b^2}

=>     c  =  \sqrt{ 10^2 -  6^2}

=>     c  = 8 \  km

6 0
4 years ago
Were the continents once joined together as a supercontinent? Give 3 pieces of evidence to support Alfred Wegeners Theory of Con
sattari [20]
Yes! Fossils, The outlines of the continents and geological features .
5 0
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Mercury was named after the roman god of speed why is it an appropriate name for the planet
elena-14-01-66 [18.8K]
Because it's the planet in our solar system with the shortest,
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5 0
3 years ago
A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t
nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 \epsilon = 1 - \dfrac{T_C}{T_H}

 0.25 = 1 - \dfrac{T_C}{464}

 T_C= (1 - 0.25) \times 464

 T_C= 0.75 \times 464

      T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

 T_C= (1 - 0.42) \times 464

 T_C= 0.58 \times 464

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8 0
3 years ago
series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s
andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

\displaystyle X_c=\frac{1}{wC}

where w is the angular frequency

w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s

Solving for C

\displaystyle C=\frac{1}{wX_c}

The reactance can be found knowing the total impedance of the circuit:

Z^2=R^2+X_c^2

Where R is the resistance, R=15 K\Omega=15000\Omega. Solving for Xc

X_c^2=Z^2-R^2

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

\displaystyle Z=\frac{V}{I}

The rms current is the peak current Ip divided by \sqrt{2}, thus

\displaystyle Z=\frac{\sqrt{2}V}{I_p}

I_p=0.65\ mA/1000=0.00065\ A

Now collect formulas

\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

Or, equivalently

\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}

X_c=36176.34\ \Omega

The capacitance is now

\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

The radius of the plates is

r=18\ cm/2=9 \ cm = 0.09 \ m

The separation between the plates is

\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

8 0
3 years ago
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