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prisoha [69]
3 years ago
11

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

r at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2.
Physics
1 answer:
ELEN [110]3 years ago
7 0

A useful formula that gives the free-fall distance from rest in 'T' seconds:

                       D = (1/2 G) x (T²)

      G = 9.81 m/s²
1/2 G = 4.905 m/s²

                       D (5 seconds) = (4.905 m/s²) x (5 sec)²

                                              = (4.905 m/s²) x (25 sec²)

                                              =        122.625 meters .

Since the tower-top is 100m above ground,
the depth of the well, to the top of the water,
accounts for the additional  22.625 meters.

My question is:  How do you know exactly when the stone hit the water ?

You probably stood at the top of the well and listened for the
sound of the 'plop'.  But it took some time after the stone hit
the water for the sound of the plop to come back up to you. 
Well, can't you just subtract that time ?  Yes, but you need
to know how much time to subtract.  That depends on the
depth of the well ... which is exactly what you're trying to
determine, so you don't know it yet.

Oh well.  That's a deep subject.
 
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A circuit consists of a 9.3-mH coil, a 16.0-V battery, a parallel combination of a 19-Ω resistor and a 6.0-Ω resistor, and a swi
Alla [95]

Answer:

τ = 0.00203 seconds

Explanation:

The time constant τ in a R-L circuit is given by

τ = L/R

First we have to find out the equivalent resistance of the circuit.

Since there is a parallel combination of 19 Ω and 6.0 Ω resistor

Req = 19*6/19+6

Req = 4.56 Ω

Now we can find out the time constant

τ = L/R

τ = 0.0093/4.56

τ = 0.00203 seconds

Therefore, the time constant of this circuit is 0.00203 seconds.

8 0
3 years ago
Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux
Fofino [41]

Answer:

a

  \phi = 1.78 *10^{-7} \  Weber

b

 L  = 1.183 *10^{-7} \  H

Explanation:

From the question we are told that

   The radius is  r = 6 \ cm =  \frac{6}{100} =  0.06 \ m

   The current it carries is  I  = 1.50 \ A

     

The  magnetic flux of the coil is mathematically represented as

       \phi = B  * A

Where  B is the  magnetic field which is mathematically represented as

         B  =  \frac{\mu_o  * I}{2 *  r}

Where  \mu_o is the magnetic field with a constant value  \mu_o  =  4\pi * 10^{-7} N/A^2

substituting  value

          B  =  \frac{4\pi * 10^{-7}   * 1.50 }{2 *  0.06}

          B  =  1.571 *10^{-5} \ T

The area A is mathematically evaluated as

       A  = \pi r ^2

substituting values

       A  = 3.142 *  (0.06)^2

       A  = 0.0113 m^2

the magnetic flux is mathematically evaluated as    

        \phi = 1.571 *10^{-5} * 0.0113

         \phi = 1.78 *10^{-7} \  Weber

The self-inductance is evaluated as

       L  =  \frac{\phi }{I}

substituting values

        L  =  \frac{1.78 *10^{-7} }{1.50 }

         L  = 1.183 *10^{-7} \  H

7 0
3 years ago
Chintu and raven build a roccket, which moved from the earth to about 86m into the shy. It tales 3.7 seconds to reach the rocket
vaieri [72.5K]

   Speed = (distance covered) / (time to cover the distance)..

From the ground to the highest point,
the rocket's AVERAGE speed is

                     (86 m) / (3.7 sec)  =  23.2 m/s  (rounded) .
5 0
3 years ago
How is velocity and instantaneous speed alike
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5 0
3 years ago
You are playing right field for the baseball team. Your team is up by one run in the botton of the last inning of the game when
VARVARA [1.3K]

Answer:

The correct response is "No". The further explanation is given below.

Explanation:

The given values are:

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\theta = 30^{\circ}

Distance

d = 20 m

Speed

s = 8.0 m/s

Now,

⇒  y(t) = \frac{1}{2}\times a\times t^2 + v0\times t\times (sin \theta)

    0 = -4.9t^2 + v0\times t\times (sin 30)

    x = v0\times (cos \theta)\times t

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    v0\times t = \frac{65}{Cos30}

    0 = -4.9t^2 + 65\times \frac{sin30}{cos30}

    t = 2.767 \ sec

So,

⇒  d = r\times t

    20 = 8\times t

      t=2.5 \ sec

Therefore, 2.5 < 2.767, so it won't get there.

6 0
3 years ago
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