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3 years ago

Mass of water= 357g density water= 1.0g/cm3

1 answer:

8 0

$m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3$

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What is the value for the kinetic energy for a n = 5 Bohr orbit electron in zeptoJoules?

Ne4ueva [31]

Answer:

The kinetic energy is 86.6 zepto joules.

Explanation:

Given that,

Number of orbit n =5

We know that,

Bohr's radius for hydrogen atom is

$r = 0.53\times10^{-10}\times n^2\ m$

Now, put the value of n in the formula of radius

$r=0.53\times10^{-10}\times5^2$

$r =1.33\times10^{-9}\ m$

We need to calculate the kinetic energy

Using formula of kinetic energy

$E_{k}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{e^2}{2\times r_{s}}$

Put the value into the formula

$E_{k}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times1.33\times10^{-9}}$

$E_{k}=8.66\times10^{-20}\ J$

We know that,

$1\ zepto\ joule=1\times10^{-21}\ J$

The kinetic energy is

$E_{k}=\dfrac{8.66\times10^{-20}}{1\times10^{-21}}$

$E_{k}=86.6\ zepto\ Joules$

Hence, The kinetic energy is 86.6 zepto joules.

8 0

3 years ago

Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.

luda_lava [24]

A )
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81 + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52

4 0

4 years ago

Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp

maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = $\dfrac{N}{C}m$

c = Speed of light = m/s

Momentum is given by

$\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s$