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3 years ago

5

A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and

the length of the chain is 1.9 m. The acceleration of gravity is 9.81 m/s 2 . Find the tension in the chain at the point where the chain is supporting the block. Answer in units of N.

1 answer:

docker41 [41]3 years ago

4 0

Answer:

T = 451.26 N

Explanation:

It is given that,

The mass of block, m = 46 kg

Mass of the chain, m' = 19 kg

Length of the chain, l = 1.9 m

Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.

$T=mg$

$T=46\ kg\times 9.81\ m/s^2$

T = 451.26 N

So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.

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Answer:

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Explanation:

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3 years ago

An electric current of 200 A is passed through a stainless steel wire having a radius R of 0.001268 m. The wire is L = 0.91 m lo

denis23 [38]

Answer:

The value is $T_m = 435.2 \ K$

Explanation:

From the question we are told that

The current is $I = 200 \ A$

The radius is $R = 0.001268 \ m$

The length of the wire is $L = 0.91 \ m$ \

The resistance is $R = 0.126 \ \Omega$

The outer surface temperature is $T _o = 422.1 \ K$

The average thermal conductivity is $\sigma = 22.5 W/mK$

Generally the heat generated in the stainless steel wire is mathematically represented as

$Q = \frac{Power}{ \pi r^2L}$

$Q = \frac{I^2 R}{ \pi r^2L}$

=> $Q = \frac{200^2 * 0.126}{3.142 * (0.001268)^2 * 0.91}$

=> $Q = 1.096*10^{9}\ W/m^3$

Generally the middle temperature is mathematically represented as

$T_m = T_o + \frac{Q * r^2 }{ 6 * \sigma }$

$T_m = 422.1 + \frac{1.096*10^{-9} * 0.001268^2}{6 * 22.5}$

$T_m = 435.2 \ K$

4 0

3 years ago

A ball is thrown vertically upwards at 19.6 m/s. For its complete trip (up and back down to the starting position), its average

stira [4]

When the ball reaches its original position, it will have the same speed (but would be traveling in the opposite direction). So the average speed is

$\dfrac{19.6\,\frac{\mathrm m}{\mathrm s}-19.6\,\frac{\mathrm m}{\mathrm s}}{\Delta t}=0$

regardless of how long the ball was in the air.

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3 years ago

At 20°C the vapor pressure of dry ice is 56.5 atm. If 10g of dry ice (solid CO2) is placed in an evacuated 0.25 L chamber at a c

jonny [76]

Answer:

Pressure of the gas is greater than the vapor pressure, therefore all the solid will not sublime.

Explanation:

To solve this problem we use ideal gas equation.

PV =nRT

If Presure of the gas (P) is greater than vapor pressure (56.5 atm), all the solid will not sublime. But If P < 56.5 atm, all of it will sublime.

where;

P is pressure of the gas in atm

V is volume of gas in Litre

T is absolute temperature of the gas in Kelvin

n is number of gas moles

R is ideal gas constant or Boltzmann constant = 0.082057 L atm K⁻¹ mol⁻¹

Given T = 20 °C = 273 + 20 = 293K

Volume = 0.25L

n = Reacting mas (m)/Molar mass(M)

Molar mass of CO₂ = 12 + (16X2) = 12+32 = 44g/mol

Reacting mass = 10g

n = m/M ⇒ 10/44

n = 0.2273

PV =nRT ⇒ P = nRT/V

P = (0.2273X0.082057 X293)/0.25

P = 21.8596 atm

Therefore, since Pressure of the gas is greater than the vapor pressure, all the solid will not sublime.

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4 years ago

When mass m is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 180 h

Katena32 [7]

The frequency of the nth-harmonic of a string is given by
$f_n = \frac{n}{2L} \sqrt{ \frac{T}{\mu} }$
where n is the number of the harmonic, L is the length of the string, T the tension and $\mu$ the linear density.

In our problem, since the mass m is tied to the string, the tension is equal to the weight of the object tied:
$T=mg$
Substituting into the first formula, we have
$f_n = \frac{n}{2L} \sqrt{ \frac{mg}{\mu} }$

In our problem we have n=2 (second harmonic). In the previous equation, the only factor which is not constant between the first and the second part of the problem is m, the mass. So, we can rewrite everything as
$f_2 = K \sqrt{m}$
where we called
$K= \frac{2}{2L} \sqrt{ \frac{g}{\mu} }$

In the first part of the problem, the mass of the object is m and $f_2 = 180 Hz$ . So we can write
$180 Hz = K \sqrt{m}$

When the mass is increased with an additional 1.2 kg, the relationship becomes
$270 Hz = K \sqrt{(m+1.2 Kg)}$

By writing K in terms of m in the first equation, and subsituting into the second one, we get
$180 Hz \sqrt{ \frac{m+1.2 Kg}{m} }=270 Hz$
and solving this, we find
$m=0.96 kg$

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4 years ago