Here current is flowing through the copper wire so this shows that copper is good conductor of electricity.
It is having less resistance as it conducts the current easily
Now a rubber coating on it will protect us from electric shock
So this property shows that rubber is a bad conductor of electricity
It is having large electrical resistance due to which it will conduct no current
Rubber : - No transmittance of electricity
copper :- good transmittance of electricity
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
The sum is the result of adding 9260 and 3240 together. Each number can
be broken down into constituent parts in order to make addition easier.
Each place in the number represents its value, so a 2 in the hundreds
place represents 200.
You can separate numbers out this way to
make it easier to add them. 9260 can be broken down into 9000+200+60
while 3240 is 3000+200+40. You can then add these six numbers together.
60+40 = 100
200+200 = 400
9000+3000 = 12000
Then add your three partial results together to receive the final answer:
12000+400+100 = 12500
Answer:
993.52 Hz
Explanation:
The frequency of sound emitted by the stationery train is 1057 Hz.
The car travels away from the train at 20.6 m/s.
The frequency the observer hears is given by the formula:
where v = velocity of sound = 343 m/s
vo = velocity of observer
f = frequency from source
This phenomenon is known as Doppler's effect.
Therefore:
The frequency heard by the observer is 993.52 Hz.
The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
(1)
The work done is equal to the elastic energy stored by the compressed spring:
where
is the spring constant and
is the compression of the spring. If we substitute the numbers, we find:
And now we can use eq.(1) to calculate the average power output:
