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2 years ago

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A stone dropped from the top of a building reaches the ground with a velocity of 49ms¹. If the acceleration due to gravity is 9.

8ms², calculate the time for which the stone is falling freely. [Answer: 5s]

1 answer:

bezimeni [28]2 years ago

6 0

Explanation:

Given:

v₀ = 0 m/s

v = 49 m/s

a = 9.8 m/s²

Find: t

v = at + v₀

49 m/s = (9.8 m/s²) t + 0 m/s

t = 5 s

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Answer:

The maximum height a person's brain could be above his heart is: 1.28 meter.

Explanation:

We need to know what is the normal blood pressure ours hearts so there is a rate: 120/80 (mmHg) and the average will be: 100 (mmHg) and using the Pascal law that relate pressure, density, gravity and height like: $P_{2} = pgh_{1} - pgh_{2} + P_{1}$ , where P is pressure, p is density, g is the gravity acceleration and h is the height. Now we can find the height and delta of pressure will be: P2-P1 = 100 (mmHg), knowing that 1(mmHg) is equal to 133 Pa, we can do the convertion to 13332.2 (Pa), now because the units of Pascal are Newton/(meter^2). Then we solve the formula to get the height: $\frac{P2-P1}{pg} =h$ so we get: $\frac{13332.2}{(1060*9,81)}=Height=1.28(meters).$

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

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A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: <span>North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.</span>

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The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

Fudgin [204]

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

$V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3$

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

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