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2 years ago

What is the difference between first aid and first aid kit

Physics

1 answer:

Archy [21]2 years ago

7 0

Answer:

One has the word kit and the other doesn't

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after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

2 years ago

Altitude is the angle measured above ____. North Pole horizon equator zenith

juin [17]

Altitude is the angle measured above the horizon

5 0

3 years ago

Read 2 more answers

If each pull-up requires 300 J and Ben does a pull-up in 2 seconds, what is his power? 150 watts 300 watts 600 watts 750 watts

leonid [27]

Answer:

150 watts

Explanation:

300/2 = 150 watts

6 0

2 years ago

Read 2 more answers

The picture below shows a wheelbarrow. Using a wheelbarrow can make it easier to lift a heavy object by

timofeeve [1]

Answer:

B IS CORRECT

Explanation:

8 0

2 years ago

To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that

noname [10]

Answer:

a. 2v₀/a b. 2v₀/a

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. = 1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)

= 2v₀/a

4 0

3 years ago