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3 years ago

A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?

Physics

1 answer:

pychu [463]3 years ago

3 0

Answer:

C. 30.6m

Explanation:

To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

$H=ut+\frac{1}{2}gt^2$

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.

Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

$H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m$

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A person takes a trip, driving with a constant speed of 99.5 km/h, except for a 26.0-min rest stop. The person's average speed i

Korvikt [17]

Answer:

1.65 h

121.39 km

Explanation:

Given that

speed of the driver, v = 99.5 km/h

time spent resting, t = 26 min

Average speed of the driver = 73.6 km/h

check attachment for calculation and how I arrived at the answer

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2 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

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3 years ago

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

bija089 [108]

This question is incomplete, the complete question is;

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.

At what rate is the magnetic field changing?

Answer:

the magnetic field changing at the rate of 9.33 m T/s

Explanation:

Given the data in the question;

Electric field E = 7 mV/m

radius r = 1.5 m

Now, from Faraday law of induction;

∫E.dl = d∅/dt

E∫dl = A( dB/dt )

E( 2πr ) = πr² ( dB/dt )

( 0.007 ) = (r/2) ( dB/dt )

( 0.007 ) = 0.75 ( dB/dt )

dB/dt = 0.007 / 0.75

dB/dt = 0.00933 T/s

dB/dt = ( 0.00933 × 1000) m T/s

dB/dt = 9.33 m T/s

Therefore, the magnetic field changing at the rate of 9.33 m T/s

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3 years ago

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Explanation:

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