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3 years ago

The prismatic bar has a cross-sectional area

Physics

1 answer:

ludmilkaskok [199]3 years ago

8 0

solution:

$sum fx=0\\
n=\frac{1}{2}\times\frac{w_{0}}{a}(2a-x)^2\\
n=\frac{w_{0}}{20}(2a-x)^2\\
angle is the normal shus\\
\sigma =\frac{N}{A}\\
\frac{\frac{w_{0}}{2a}(2a-x)^2}{A}\\
\sigma(x)=\frac{w_{0}}{2aA}[2a-x]^2$

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Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k

IrinaVladis [17]

Answer:

magnitude = 3

unit vector = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

Explanation:

Given vectors:

u = 2i + 2j - k

v = -i + k = -i + 0j + k

(a) u x v is the cross product of u and v, and is given by;

$u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]$

u x v = i(2+0) - j(2 - 1) + k(0 - 2)

u x v = 2i - j - 2k

Now the magnitude of u x v is calculated as follows:

| u x v | = $\sqrt{2^2 + (-1)^2 + (-2)^2}$

| u x v | = $\sqrt{4 + 1 + 4}$

| u x v | = $\sqrt{9}$

| u x v | = 3

Therefore, the magnitude of u x v is 3

(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e

û = $\frac{u X v}{|u X v|}$

u x v = 2i - j - 2k [<em>calculated in (a) above</em>]

|u x v| = 3 [<em>calculated in (a) above</em>]

∴ û = $\frac{2i - j - 2k}{3}$

∴ û = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

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3 years ago

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3 years ago

Batteries are rated in terms of ampere-hours (A·h). For example, a battery that can produce a current of 2.00 A for 3.00 h is ra

Reptile [31]

(a) 423 J

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$P=\frac{E}{t}$

However, the power is also the product of the voltage (V) and the current (I):

$P=VI$

Linking the two equations together,

$\frac{E}{t}=VI\\E=VIt$

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V = 9.0 V

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The battery has a total energy of E = 423 J. (2)

1 Watt (W) is equal to 1 Joule (J) per second (s):

$1 W = \frac{1 J}{1 s}$

so 1 kW corresponds to 1000 J/s:

$1 kW = \frac{1000 J}{1 s}$

Multiplying both side by 1 hour (1 h):

$1 kW \cdot h = \frac{1000 J}{1 s} 1 h$

and $1 h = 3600 s$ , so

$1 kWh = \frac{1000 J}{1 s}\cdot 3600 s =3.6\cdot 10^6 J$

So we find the conversion between kWh and Joules. So now we can convert the energy from Joules (2) into kWh:

$1 kWh = 3.6\cdot 10^6 J = x : 423 J\\x=\frac{1 kWh \cdot 423 J}{3.6\cdot 10^6 J}=1.18\cdot 10^{-4}kWh$

And since the cost is $0.0660 per kilowatt-hour, the total cost will be

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4 years ago

Two diamonds begin a free fall from rest and from the same height 1 second apart.

Bumek [7]

Answer:

Explanation:

let t be time since the second diamond is released

a) y = ½g(t + 1)²

b) y' = ½g(t)²

25 = ½g(t + 1)² - ½gt²

25 = ½g(t² + 2t + 1) - ½gt²

25 = ½gt² + gt + ½g - ½gt²

25 = g(t + ½)

t + ½ = 25 / g

t = (25 / g) - ½

t = (25 / 9.8) - ½

t = 2.05102... ≈ 2.1 s

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3 years ago