To solve this problem divide 60 by 4.6
The answer to this problem is 13 seconds.
Answer:
I= 3.5 amps
Explanation:
Step one:
given data
rating of resistor R= 8 ohms
power P= 100W
Required
The current I
Step two
Yet this power is also given by
![P = I^2R](https://tex.z-dn.net/?f=P%20%3D%20I%5E2R)
make I subject of the formula we have
![I= \sqrt{\frac{P}{R} }](https://tex.z-dn.net/?f=I%3D%20%5Csqrt%7B%5Cfrac%7BP%7D%7BR%7D%20%7D)
substitute
![I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps](https://tex.z-dn.net/?f=I%3D%20%5Csqrt%7B%5Cfrac%7B100%7D%7B8%7D%20%7D%5C%5C%5C%5CI%3D%5Csqrt%7B12.5%7D%5C%5C%5C%5CI%3D%203.5%20amps)
And.. where is the rest of the question?
Answer:
g = 11.2 m/s²
Explanation:
First, we will calculate the time period of the pendulum:
![T = \frac{t}{n}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7Bt%7D%7Bn%7D)
where,
T = Time period = ?
t = time taken = 135 s
n = no. of swings in given time = 98
Therefore,
![T = \frac{135\ s}{98}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B135%5C%20s%7D%7B98%7D)
T = 1.38 s
Now, we utilize the second formula for the time period of the simple pendulum, given as follows:
![T = 2\pi \sqrt{\frac{l}{g}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D)
where,
l = length of pendulum = 54 cm = 0.54 m
g = acceleration due to gravity on the planet = ?
Therefore,
![(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}](https://tex.z-dn.net/?f=%281.38%5C%20s%29%5E2%20%3D%204%5Cpi%5E2%28%5Cfrac%7B0.54%5C%20m%7D%7Bg%7D%20%29%5C%5C%5C%5Cg%20%3D%20%5Cfrac%7B4%5Cpi%5E2%280.54%5C%20m%29%7D%7B%281.38%5C%20s%29%5E2%7D)
<u>g = 11.2 m/s²</u>
Answer:
The depth of the water at this point is 0.938 m.
Explanation:
Given that,
At one point
Wide= 16.0 m
Deep = 3.8 m
Water flow = 2.8 cm/s
At a second point downstream
Width of canal = 16.5 m
Water flow = 11.0 cm/s
We need to calculate the depth
Using Bernoulli theorem
![A_{1}V_{1}=A_{2}V_{2}](https://tex.z-dn.net/?f=A_%7B1%7DV_%7B1%7D%3DA_%7B2%7DV_%7B2%7D)
Put the value into the formula
![16.0\times3.8\times2.8=16.5\times x\times 11.0](https://tex.z-dn.net/?f=16.0%5Ctimes3.8%5Ctimes2.8%3D16.5%5Ctimes%20x%5Ctimes%2011.0)
![x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B16.0%5Ctimes3.8%5Ctimes2.8%7D%7B16.5%5Ctimes11.0%7D)
![x=0.938\ m](https://tex.z-dn.net/?f=x%3D0.938%5C%20m)
Hence, The depth of the water at this point is 0.938 m.