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3 years ago

What is an inertial reference frame?

Physics

1 answer:

Gennadij [26K]3 years ago

8 0

<h2>Answer: A system in which Newton's Laws are fulfilled</h2>

An inertial reference system is a reference system in which the principle of inertia is fulfilled, which is one of Newton's laws:

<em>"For a body to have acceleration, an external force must act on it" </em>

In addition, the other Newton's laws of movement are fulfilled.

Therefore, the variation of the linear momentum of the system is equal to the actual forces on the system.

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For a centrifugal compressor, the flow at the exit of the blade (state 2) has a velocity of 250 m/s with an angle of 15 degrees

shtirl [24]

The velocity at the end of vaneless space(state 3) is 965.92 m/s

<h3>What is centrifugal compressor?</h3>

It increases kinetic energy to the airstream using a rotating element and then converts it into potential energy in the form of pressure.

Temperature at state 2, T₂ =T₀ + c/2Cp

Substitute T₀ =450K, c=250m/s, Cp =1005, we get

T₂ =418.90K

From the velocity triangle, sinβ₂ =c₂/v₂

v₂ = 250/sin (90°-75°) = 965.92 m/s

Thus, the velocity at the end of vaneless space is 965.92 m/s.

Learn more about centrifugal compressor.

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6 0

3 years ago

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it

soldi70 [24.7K]

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$ .

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$ .

7 0

3 years ago

What is the term for the matter through which a mechanical wave travels?

pav-90 [236]

The answer is “ The medium”

7 0

3 years ago

will the value of g be affected by the radius of the earth? consider the real shape of the earth. compare the acceleration due t

sp2606 [1]

Answer:

Yes, the value of g affected by the radius.

Explanation:

The formula for the force of gravity of 2 objects is

$F_{gravity} = G\frac{m_{1}m_{2}}{r^2}$ , where m1 and m2 are the masses of the 2 objects, r is the radius, and G is the gravitational constant, which is approximately $6.67 \cdot 10^{-11}$ .

Therefore, as the radius if bigger, the force of gravity is going to be smaller exponentially.

5 0

4 years ago