(A) We can solve the problem by using Ohm's law, which states:
![V=IR](https://tex.z-dn.net/?f=V%3DIR)
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
![V=220 V](https://tex.z-dn.net/?f=V%3D220%20V)
and
![R=220 \Omega](https://tex.z-dn.net/?f=R%3D220%20%5COmega)
, therefore we can rearrange Ohm's law to find the current through the device:
![I= \frac{V}{R}= \frac{220 V}{220 \Omega}=1 A](https://tex.z-dn.net/?f=I%3D%20%5Cfrac%7BV%7D%7BR%7D%3D%20%5Cfrac%7B220%20V%7D%7B220%20%5COmega%7D%3D1%20A%20%20)
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
![R= \rho \frac{L}{A}](https://tex.z-dn.net/?f=R%3D%20%5Crho%20%5Cfrac%7BL%7D%7BA%7D%20)
where
![\rho](https://tex.z-dn.net/?f=%5Crho)
is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325
![T_1=25^{\circ}\approx 298 K](https://tex.z-dn.net/?f=T_1%3D25%5E%7B%5Ccirc%7D%5Capprox%20298%20K)
![V=0.025 m^3](https://tex.z-dn.net/?f=V%3D0.025%20m%5E3)
![T_2=50^{\circ}\approx 323 K](https://tex.z-dn.net/?f=T_2%3D50%5E%7B%5Ccirc%7D%5Capprox%20323%20K)
as the volume remains constant therefore
![\frac{P_1}{T_1}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
![\frac{311.325}{298}=\frac{P_2}{323}](https://tex.z-dn.net/?f=%5Cfrac%7B311.325%7D%7B298%7D%3D%5Cfrac%7BP_2%7D%7B323%7D)
![P_2=337.44 KPa](https://tex.z-dn.net/?f=P_2%3D337.44%20KPa)
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass ![m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}](https://tex.z-dn.net/?f=m_1%3D%5Cfrac%7BP_1V%7D%7BRT_1%7D%3D%5Cfrac%7B311.325%5Ctimes%200.025%7D%7B0.0287%5Ctimes%20298%7D)
![m_1=0.91 kg](https://tex.z-dn.net/?f=m_1%3D0.91%20kg)
Final mass ![m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}](https://tex.z-dn.net/?f=m_2%3D%5Cfrac%7BP_2V%7D%7BRT_2%7D%3D%5Cfrac%7B311.325%5Ctimes%200.025%7D%7B0.0287%5Ctimes%20323%7D)
![m_2=0.839](https://tex.z-dn.net/?f=m_2%3D0.839)
Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
When the velocity of an object changes, it is acted upon by a force
Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see.
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car
zone 6 = area directly behind your car.
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