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natta225 [31]
3 years ago
9

What is the purpose of cpr??

Physics
2 answers:
Ket [755]3 years ago
8 0
Its main purpose is to restore partial flow of oxygenated blood to the brain and heart in a short amount of time.
Marta_Voda [28]3 years ago
7 0

Answer:

To restore a heart beat

Explanation:

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A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti
Korvikt [17]

Answer:

The strength of the electric field is 1.35\times10^{4}\ N/C.

Explanation:

Given that,

Speed v= 5.05\times10^{5}\ m/s

Time t= 3.90\times10^{-7}\ s

Angle = 45°

We need to calculate the acceleration

Using equation of motion

v = u+at

5.05\times10^{5}=0+a\times3.90\times10^{-7}

a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}

a=1.29\times10^{12}\ m/s^2

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

F= ma=qE

ma = qE

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}

E=1.35\times10^{4}\ N/C

Hence, The strength of the electric field is 1.35\times10^{4}\ N/C.

3 0
3 years ago
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d
Archy [21]

Answer:

The voltage is   V =   0.993V_b

Explanation:

From the question we are told that

   The time that has passed is  t = \frac{\tau}{2}

 Here \tau is know as the time constant

    The voltage of the  power source is   V_b

Generally the voltage equation for charging a capacitor is mathematically represented as

       V =  V_b  [1 - e^{- \frac{t}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]

=>   V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }]

=>   V =  V_b  [1 - e^{- \frac{1}{2} }]

=>   V =   0.993V_b    

5 0
3 years ago
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the
asambeis [7]
EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
7 0
3 years ago
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If you had a planet to choose from to live in what would it be?
postnew [5]
I personally would live on Mars cuz that is red cuz
4 0
3 years ago
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