(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:![6.86 m/s^2](https://tex.z-dn.net/?f=6.86%20m%2Fs%5E2)
Explanation:
Given
mas of car=870 kg
coffee mug mass=0.47 kg
coefficient of static friction between mug and roof ![\mu _s= 0.7](https://tex.z-dn.net/?f=%5Cmu%20_s%3D%200.7)
Coefficient of kinetic Friction ![\mu _k=0.5](https://tex.z-dn.net/?f=%5Cmu%20_k%3D0.5)
maximum car acceleration is ![\mu \times g](https://tex.z-dn.net/?f=%5Cmu%20%5Ctimes%20g)
here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect
![a_{max}=0.7\times 9.8=6.86 m/s^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%3D0.7%5Ctimes%209.8%3D6.86%20m%2Fs%5E2)
Answer:
t = 2.2 s
Explanation:
Given that,
A person observes a firework display for A safe distance of 0.750 km.
d = 750 m
The speed of sound in air, v = 340 m/s
We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.
![v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%7D%7Bt%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7Bd%7D%7Bv%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B750%5C%20m%7D%7B340%5C%20m%2Fs%7D%5C%5C%5C%5Ct%3D2.2%5C%20s)
So, the required time is 2.2 seconds.
Least count of the pulse stopwatch is given by
![\Delta t = 0.1 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%200.1%20s)
this means each division of the stopwatch will measure 0.1 s of time
After 3 journeys from one end to other we can see that total time that is measured here is shown by the clock as 52nd division
So here total time is given as
Time = (Number of division) (Least count)
now we will have
![T = 52 \times 0.1s](https://tex.z-dn.net/?f=T%20%3D%2052%20%5Ctimes%200.1s)
![T = 5.2 s](https://tex.z-dn.net/?f=T%20%3D%205.2%20s)