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bulgar [2K]
3 years ago
8

A uniform brick of length 18 m is placed

Physics
1 answer:
yuradex [85]3 years ago
3 0

Answer:

13.5

Explanation:

x= Length

.5 = Center Point

x=0.5(9)= 4.5

So total length over the edge will be given as

Length (L) =9+4.5=13.5

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Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressur
Natalka [10]

Answer:  a. 17.7 KJ/Mol

b. T=210K

Explanation:

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressure is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate.

the question isnt completely originally, but we could look at the likely derivation from the questions

(a) the standard enthalpy of vaporization

using the clausius clapeyron equation

In (PT1vap / PT2vap) = delta H (vap) / R ( (1/T1) - (1/T2) )

In (35Torr/253Torr) = delta H (vap) / 8.3145 ( (1/189.55) - (1/161.2) )

Therefore, Delta H (vap) = 17.7 KJ/Mol

b. Also the boiling point

What is the normal boiling point of arsine?

At the boiling point Pvap = atmospheric pressure = 1 atm=760 torr

substitution into the equation as stated in question 1

ln(760/253)=17700/8.314(1/189.55-1/T)

T=210K

5 0
3 years ago
A machinist is required to manufacture a circular metal disk with area 1300 cm2. (a) What radius produces such a disk
vichka [17]

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

6 0
3 years ago
4
telo118 [61]
Ik what it is hit me up for it
5 0
3 years ago
Which layer of the sun's atmosphere are you looking at when you look at me in image of the Sun
Hitman42 [59]
The answer is Photosphere.  The photosphere is the lowest layer<span> of the </span>solar<span> atmosphere. It is essentially the </span>solar<span> "surface" that </span>we see<span> when </span>we look<span> at the </span>Sun in "white" light.  It is <span>like a glowing fog, so at a distance, it </span>looks<span> solid, the same way a cloud looks solid from a distance.</span>
7 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
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