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3 years ago

9

In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

corner, the pressure is 5.48 atm. Calculate the deflection angle of the corner.

1 answer:

yulyashka [42]3 years ago

4 0

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

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The mass of a basketball is three times greater than the mass of a softball. Compare the momentums of a softball and a basketbal

ki77a [65]

Answer:

The momentum of the basketball is three times that of the softball. Momentum equals mass times velocity. Therefore, if the basketball and softball are moving at the same velocity, and the basketball has three times the mass of the softball, the basketball has three times the momentum of the softball.

Explanation:

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3 years ago

2. Two people, one on Earth and the other on the Moon, try and lift 200 kg blocks. Which person

Art [367]

Answer:

The person on Earth will have to exert more force to lift their block

Explanation:

The mass of the blocks to be lifted = 200 kg

The location of the first person = On Earth

The location of the second person = On the Moon

The force a person will have to exert to lift the block = The weight of the block = The gravitational force, F, on the block which is given as follows;

$F = \dfrac{G \times M_1}{R^2} \times m_2$

Where;

$\dfrac{G \times M_1}{R^2}$ = The acceleration due to gravity on the Earth or the Moon, depending on the location of the block

m₂ = The mass of the block

Therefore, given that the acceleration due to gravity on the Earth is larger than the acceleration due to gravity on the Moon, the weight of the block on the Earth is larger than the weight of the block on the Moon, and the person on Earth have to exert more force to lift the heavier weight of the block on Earth than the person on the Moon will have to exert to lift the same block as the block has a lower weight on the Moon due to lower acceleration due to gravity on the Moon.

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3 years ago

What do hurricanes, tornadoes, and thunderstorms have in common?

amm1812

Answer:

low pressure systems

Explanation:

with a low pressure system over the area the storm can't form

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3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

4 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago