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3 years ago

I need help with 6a, 6b, and 6c !!

Physics

1 answer:

andrey2020 [161]3 years ago

8 0

Answer

help me too

Explanation:

have this same problem, please help

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PLEASE ANSWER ILL GIVE BRAIN!!!

zimovet [89]

What’s the question?

3 0

3 years ago

Read 2 more answers

(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,

krek1111 [17]

Answer:

$V_2 = 45.44m/s$

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

$P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K$

When we have all this values in SI apply a Energy Balance Equation,

$\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Solving for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the table of gas properties we calculate for $T_1 = 216K$ and $T_2 = 250K$

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

For T_2;

$h_2 = 250.05kJ/kg$

Substituting in equation for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s$

4 0

3 years ago

Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc

svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

$v=v_{i}+gt\\$

Also get the equation of displacement

$S=v_{i}t+(1/2)gt^{2}$

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

$v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\ And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m$

For t₂

$v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\ And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m$

For t₃

$v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\ And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m$

For t₄

$v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\ And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m$

For t₅

$v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\ And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m$

4 0

4 years ago

what’s 55mph to km/min? can someone explain to to me with the work so i can understand how to solve this

natali 33 [55]

Answer:

55 miles / hour = 1.475232 kilometers / minute

6 0

3 years ago

Read 2 more answers

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q