Explanation:
A basic law of magnetism is that unlike poles attract each other. Two bar magnets can illustrate this. ... A pole of the second is brought, in turn, near each of the two ends of the hanging magnet.
Second Law of Magnetism
The force is in direct proportion to the product of the forces of the pole. The force exists in inverse proportion to the square of the middle distance between the poles. The force is dependent on the specific medium in which the magnets are placed.
Answer:
A. 46.15 ohms
B. 1248 W
C. 2.25×10⁶ J
Explanation:
A. Determination of the resistance.
Current (I) = 5.2 A
Voltage (V) = 240 V
Resistance (R) =?
From ohm's law,
V = IR
240 = 5.2 × R
Divide both side by 5.2
R = 240 / 5.2
R = 46.15 ohms
B. Determination of the power.
Current (I) = 5.2 A
Voltage (V) = 240 V
Power (P) =?
P = IV
P = 240 × 5.2
P = 1248 W
C. Determination of the energy.
Power (P) = 1248 W
Time (t) = 30 minutes
Energy (E) =?
Next, we shall convert 30 minutes to seconds. This can be obtained as follow:
1 min = 60 s
Therefore,
30 min = 30 min × 60 s / 1 min
30 min = 1800 s
Finally, we shall determine the energy. This can be obtained as illustrated below:
Power (P) = 1248 W
Time (t) = 1800 s
Energy (E) =?
E = Pt
E = 1248 × 1800
E = 2.25×10⁶ J
Answer:
The velocity is ![v = 8.743 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%208.743%20%5C%20m%2Fs)
Explanation:
From the question we are told that
The frequency of the signal sent out is ![f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz](https://tex.z-dn.net/?f=f_s%20%20%3D%20%2038.0%20%5C%20kHz%20%20%3D%20%2038.0%20%2A10%5E%7B3%7D%20%5C%20Hz)
The frequency of the signal received is ![f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz](https://tex.z-dn.net/?f=f_r%20%20%3D%20%2040.0%20%5C%20kHz%20%20%3D%20%2040.0%20%2A10%5E%7B3%7D%20%5C%20Hz)
The speed of sound is ![v_s = 341 \ m/s](https://tex.z-dn.net/?f=v_s%20%20%3D%20%20341%20%5C%20m%2Fs)
Generally the frequency of the sound received is mathematically represented as
![f_r = f_s [\frac{v_s + v}{v_s - v} ]](https://tex.z-dn.net/?f=f_r%20%3D%20%20f_s%20%20%5B%5Cfrac%7Bv_s%20%20%2B%20v%7D%7Bv_s%20%20-%20v%7D%20%5D)
where v is the velocity of the object
=> ![40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ]](https://tex.z-dn.net/?f=40%20%2A10%5E%7B3%7D%20%3D%20%2038%20%2A10%5E%7B3%7D%20%2A%20%20%20%5B%5Cfrac%7B341%20%20%2B%20v%7D%7B341%20%20-%20v%7D%20%5D)
=> ![1.05263 = \frac{341+v }{341-v}](https://tex.z-dn.net/?f=1.05263%20%3D%20%5Cfrac%7B341%2Bv%20%7D%7B341-v%7D)
=> ![358.94 - 1.05263v = 341 + v](https://tex.z-dn.net/?f=358.94%20-%201.05263v%20%20%3D%20%20341%20%2B%20v)
=> ![17.947 = 2.05263 v](https://tex.z-dn.net/?f=17.947%20%3D%20%202.05263%20v)
=> ![v = 8.743 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%208.743%20%5C%20m%2Fs)
Answer:
Explanation:
a ) Velocity vs time graph will be a straight line with negative slope , making a slope of 2 because slope of v-t graph gives the value of acceleration.
b ) v = 0 , u = 28 m/s
t = ?
v = u - at
0 = 28 - 2 t
t = 14 s
shuttle velocity will become zero after 14 sec.
c ) v ² = u² - 2 a s
0 = 28² - 2 x 2 x s
s = 28² / 4
= 196 m .
If you know the height of the ship while in calm water,you can subtract the height of the ship with aware at its peak point and it will tell yoi how tall the wave is.