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A 2.2-kg model rocket is launched vertically and reaches an altitude of 70 m with a speed of 30 m/s at the end of powered flight

kirill115 [55]

Answer:

$r_b= (30.8\hat{i} + 69.96 \hat{j}) m$

Explanation:

given,

mass = 2.2 kg

altitude(r₀) = (70 j) m

speed = 30 m/s

m_a = 0.77 kg

m_b =1.43 kg

part A strike ground (r_a)= (80 i) m

t = 6 s

$r = r_0 + v_ot-\dfrac{1}{2}gt^2$

$r = 60\hat{j} + (30\hat{j})\times 6-\dfrac{1}{2}\times 9.8 \times 6^2$

r = 63.6 j m

by conservation of energy

$mr = m_ar_a+m_br_b$

$2.2\times 63.6\hat{j} = 0.77\times (-80 \hat{i})+2\times r_b$

$r_b= (30.8\hat{i} + 69.96 \hat{j}) m$

8 0

3 years ago

Determine the magnitude of the resultant force and the moment about the origin. Note: the symbol near the 140 N-m moment are not

arlik [135]

Answer:

R = 148.346 N

M₀ = - 237.2792 N-m

Explanation:

Point O is selected as a convenient reference point for the force-couple system which is to represent the given system

We can apply

∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N

∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N

Then

R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)

⇒ R = 148.346 N

Now, we obtain the moment about the origin as follows

M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)

We can see the pic shown in order to understand the question.

7 0

3 years ago

The minimum recommended standards for the operating system, processor, primary memory (RAM), and storage capacity for certain so

nlexa [21]

Answer:System requirements

Explanation:

its right.

8 0

2 years ago

Read 2 more answers

Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th

Archy [21]

Answer:

The turbine produces 955.53 KW power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

P = 955.53 KW

5 0

3 years ago

Do the coil resistances have any effect on the plots?

PolarNik [594]

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.

8 0

2 years ago