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zzz [600]
2 years ago
6

A motorboat embarks on a trip, heading downstream in a river in which the current flows at a rate of 1.5m/s. After 30.0 minutes,

the boat has traveled a distance of 24.3 km downstream. How long will it take the boat to travel upstream to its original point of embarkation
Physics
1 answer:
Natali5045456 [20]2 years ago
3 0

Answer:

t=2413s

Explanation:

From the question we are told that:

Velocity v=1.5m/s

Time t=30min=>30*60=>1800

Distance d=24.3km

Generally the Newton's equation for Speed going down the stream is mathematically given by

 v + u = \frac{d}{t}

 1.5+v=frac{24300}{1800}

 v=12m/s

Therefore

 v + u = \frac{d}{t}

 t=\frac{24300}{12-1.5}

 t=2413s

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-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
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An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.
7 0
2 years ago
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various
Schach [20]

Answer:

q  =  -461532.5 \ C

Explanation:

From the question we are told that

     The  electric filed is  E  =  102 \ N/C  

Generally according to Gauss law

=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

    - E  A  =  \frac{q}{\epsilon_o }

Here   q is the excess charge on the surface of the earth

          A is the surface  area of the of the earth which is mathematically represented as

     A  =  4\pi r^2

Where r is the radius of the earth which has a value r = 6.3781*10^6 m

 substituting values

    A  = 4 * 3.142  *   (6.3781*10^6 \ m)^2

    A  =5.1128 *10^{14} \ m^2

So

   q  =  -E  * A  *  \epsilon _o

Here \epsilon_o s the permitivity of free space with value

          \epsilon_o  =  8.85*10^{-12} \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

     q  =  -102  * 5.1128 *10^{14}  *  8.85 *10^{-12}

     q  =  -461532.5 \ C

6 0
3 years ago
3. A cat pushes a 0.25-kg toy with a net force of 8 N. According to Newton's second
jek_recluse [69]
  • Mass=0.25kg
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\\ \sf{:}\!\implies F=ma

\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}

\\ \sf{:}\!\implies Acceleration=\dfrac{8}{0.25}

\\ \sf{:}\!\implies Acceleration=32m/s^2

5 0
3 years ago
The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.
Eva8 [605]

Answer: 29.17m/s^2

Explanation:

Given the following :

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v = u + at

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Plugging our values

525 = 0 + a × 18

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8 0
3 years ago
I need help with question two
zaharov [31]
I can't see numbers here, so here are all the answers:
1) the frequency is c/λ = 3e8/556e-9 = 5.39e14Hz
2) light travels at <span>299,792,458 m/s.  So in nanoseconds it's 0.299792458m.  This is about 1/3 of a meter which is about one foot.
3) length is L = ct = (</span>299,792,458 m/s)(6e-15) = 1.799e-6m or 1.799μm

6 0
2 years ago
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