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3 years ago

Is the battery sends a current of 10 A through a circuit for one hour, how many coulombs will flow through the circuit?

1 answer:

7 0

1 Ampere = 1 coulomb/second

1 hour = 3,600 seconds

1 coulomb/second = 3,600 coulomb/hour

10 Amperes x 1 hour = 36,000 coulomb/hour

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I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

3 years ago

What is the physical quantity in which force is divide by area called

pogonyaev

Explanation:

Pressure is the physical quantity having formula force per unit area.

5 0

3 years ago

A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

Ierofanga [76]

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter $a=1.6m/sec^2$

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So $F=70\times 1.6=112 N$

So external force will be 112 N

6 0

3 years ago

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

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3 years ago

Describe a perfect day

kvv77 [185]

Answer:

A bright and sunny day not worrying about work or school no family drama just a day you can relax and be yourself surrounded by the people you love.

hope this helps

have a good day :)

Explanation:

7 0

2 years ago

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