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Mice21 [21]
3 years ago
10

Two sides of a parallelogram are in the ratio 5 : 3. If its perimeter is 64 cm, find the lengths of its sides.​

Physics
1 answer:
Xelga [282]3 years ago
4 0

Answer: 32

Explanation:

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Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil
tatyana61 [14]

Answer:

The time taken by missile's clock is 4.6\times 10^{6} s

Solution:

As per the question:

Speed of the missile, v_{m = 6.5\times 10^{3}} m/s

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}         (1)

Using binomial theorem in the above eqn:

We know that:

(1 + x)^{a} = 1 + ax

Thus eqn (1) becomes:

1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}

T = \frac{2c^{2}}{v_{m}^{2}}

Now, putting appropriate values in the above eqn:

T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}

T = 4.6\times 10^{6} s

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3 years ago
If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?
qaws [65]
The answer would be erin out of all of them thank me later :)
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3 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

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3 years ago
Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt
Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

= \frac{4}{3}\pi(6400+ 50)^3 -  \frac{4}{3}\pi (6400)
^3\\\\=  \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3

Hence the volume of atmosphere is 2.6\times 10^{19} m^3

(b)

Write the ideal gas equation as foll ows:

PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles

no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules


Hence the required molecules is 13.3\times10^{43} molecules


(c)

Write the ideal gas equation as follows:

PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles

no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules

Hence the required molecules in Caesar breath is 1.2\times 10^{23} molecules

(d)

Volume fraction in Caesar last breath is as follows:  

Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}

(e)

Since the volume capacity of the human body is 500 mL.

Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath

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3 years ago
The most common form of angina is _______ angina. A. microvascular B. variant C. stable D. unstable
Paraphin [41]

The most common form of angina is stable angina.

C. stable

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2 years ago
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