The range of all electromagnetic radiation is known as the

Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.

Mars2501 [29]

Answer:

Rms speed of the particle will be $38.68\times 10^8m/sec$

Explanation:

We have given mass of the air particle $m=5\times 10^{-16}kg$

Gas constant R = 8.314 J/mol-K

Temperature is given T = $27^{\circ}C=273+27=300K$

We have to find the root mean square speed of the particle

Which is given by $v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec$

So rms speed of the particle will be $38.68\times 10^8m/sec$

5 0

3 years ago

• Can you explain why scientific theories will change over time?

marishachu [46]

Because as we develope into our times from today to tomorrow and the next, us as a human race will never stop discovering

3 0

3 years ago

Read 2 more answers

A bullet of mass m = 40~\text{g}m=40 g, moving horizontally with speed vv, strikes a clay block of mass M = 1.35M=1.35 kg that i

Sveta_85 [38]

Answer:

v > 133.5 m/s

Explanation:

Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.

Let's start by using the concepts of energy to find the velocity at the top of the circle

Initial. Top circle

Em₀ = K + U = ½ m v² + m g y

If we place the reference system at the bottom of the cycle y = 2R = L

Em₀ = ½ m v² + m g y

final. Low circle

$Em_{f}$ = K = ½ m v₁²

Emo = $Em_{f}$

½ m v² + m g y = 1/2 m v₁²

v₁² = v² + (2g L)

v₁² = v² + 2 g L

The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point

v₁ = √2g L

We already have the speed system at the bottom we can use the moment

Starting point before crashing

p₀ = m v₀

End point after collision at the bottom of the circle

$p_{f}$ = (m + M) v₁

The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved

p₀ = $p_{f}$

m v₀ = (m + M) v₁

v₀ = (m + M) / m v₁

Let's replace

v₀ = (1+ M / m) √ 2g L

Let's reduce to the SI system

m = 40 g (kg / 1000g) = 0.040 kg

Let's calculate

v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)

v₀ = 34.75 3.8417

v₀ = 133.5 m / s

the velocity must be greater than this value

v > 133.5 m/s

4 0

3 years ago

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, the

ICE Princess25 [194]

Complete Image

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by $\Delta L_{thick}$ and $\Delta L_{thin}$ , respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e $\frac{1}{2}$ option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

$E = \frac{\sigma}{\epsilon}$

Where $\sigma$ is the stress = $\frac{Force}{Area}$

$\epsilon$ is the strain = $\frac{\Delta L}{L}$

Making Strain the subject

$\epsilon = \frac{\sigma}{E}$

now in this question we are that the same tension was applied to both wires so

$\frac{\sigma}{E}$ would be constant

Hence

$\frac{\Delta L}{L} = constant$

for the two wire we have that

$\frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}$

Looking at young modulus formula

$E = \frac{\frac{F}{A} }{\frac{\Delta L}{L} }$

$E * \frac{\Delta L }{L} = \frac{F}{A}$

$A * \frac{\Delta L}{L} = \frac{F}{E}$

Now we are told that a comprehensive force is applied to the wire so for this question

$\frac{F}{E}$ is constant

And given that the length are the same

so

$A_1 \frac{\Delta L_{thin}}{L_{thin}} = A_2 \frac{\Delta L_{thick}}{L_{thick}}$

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

Assuming that

$A_2 = 2 A_1$

So

$A_1 \frac{\Delta L_{thin}}{L_{thin}} = 2 A_1 \frac{\Delta L_{thick}}{L_{thick}}$

$\frac{\Delta L_{thin}}{L_{thin}} = 2 \frac{\Delta L_{thick}}{L_{thick}}$

From the question the length are equal

$\Delta L_{thin} =2 \Delta L_{thick}$

So

$\frac{\Delta L_{thick}}{\Delta L_{thin} } = \frac{1}{2}$

Hence the ratio is less than 1

6 0

3 years ago

A Carnot engine operates between two heat reservoirs at temperatures TH and TC. An inventor proposes to increase the efficiency

Dovator [93]

Answer:

A) Efficiency = 1 - (TC/TH)

B) Comparing this with the efficiency of the original carnot engine, the efficiency is the same

Explanation:

The formula for efficiency of an original carnot engine is;

e = 1 - T(C) /T(H) ——— eq 1

In like manner, for a composite engine, the efficiency is;

e(12) = (W1 + W2)/Q(H1)

Where W1 is work done by 1st engine; W2 is work done by second engine and Q(H1) is the heat input to the first engine.

Now the total work done is;

W = Q(H) + Q(C)

Where Q(H) is the heat input and Q(C) is the heat released.

Thus,

e(12) = [Q(H1) + Q(C1) + Q(H2) + Q(C2)] / Q(H1)

Now, from the earlier e(12) equation compared to this, QH2 = -QC1

Thus;

e(12) = [Q(H1) + Q(C1) - Q(C1) + Q(C2)] / Q(H1)

So e(12) = [Q(H1) + Q(C2)] / Q(H1)

So e(12) = 1 + [Q(C2)/Q(H1)] ———eq 2

Also,

Q(C2) /Q(H2) = (-Tc/T')

Where T' is intermediate temperature.

So, simplifying that,

Q(C2) = -Q(H2) (Tc/T')

This is also equal to Q(C1) (TC/T')

But Q(C1) is also equal to;

-Q(H1) (T'/TH)

Thus; Q(C2) is now written as;

Q(C2) = -Q(H1) (T'/TH)(TC/T')

So T' will cancel out to remain;

Q(C2) = -Q(H1)(TC/TH)

Replacing this with Q(C2) in eq 2 to obtain;

e(12) = 1 + [[-Q(H1)(TC/TH)] /Q(H1)]

e(12) = 1 - TC/TH

8 0

3 years ago