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kotegsom [21]
3 years ago
6

What is the approximate number of breaths a person takes over a period of 70 years?

Physics
1 answer:
lyudmila [28]3 years ago
7 0
The approximate breathes an average person takes over 70 years is <span>588,672,000

</span>
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Why do you choose MCB in place of a fuse?
astraxan [27]

It protects the electrical appliance and the person from electrical shocks/faults.

Hope it Helped!

3 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
Read 2 more answers
Assertion(A):The distance moved by an object in unit time is called its speed. Reason (R):Faster vehicles have higher speeds. i)
ikadub [295]

Answer:

it's ii) R is correct while A is incorrect

Explanation:

cuz, both statements are correct but the reason is not the correct reason for the assertion,

8 0
3 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height
Ahat [919]

Answer:

V_{3.01}=-93.2m/s

V_{3.005}=-93.1m/s

V_{3.002}=-93.04m/s

V_{3.001}=-93.02m/s

V_{3}=-93m/s

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

Y_{3}=-99m/s

Y_{3.01}=-99.932m/s

Y_{3.005}=-99.4655m/s

Y_{3.002}=-99.18608m/s

Y_{3.001}=-99.09302m/s

Replacing these values into the formula for average velocity:

V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s

V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s

V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s

V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s

To know the actual velocity, we derive the position and we get:

V=27-40t = -93m/s

5 0
4 years ago
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