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il63 [147K]
2 years ago
13

Special relativity can be used to study an object in which frame of reference?

Physics
2 answers:
melamori03 [73]2 years ago
4 0

Answer:

B;A frame reference that has no change in velocity

Explanation:

A P E X

Tresset [83]2 years ago
3 0
Special relativity led the path for general relativity; special relativity is in a sense a special application of the rules of general relativity. While general relativity is in position to tackle all of these problems, special relativity can tackle only problems in inertial frames. Inertial frame means that the frame of reference is inot accelerating. So, we disqualify answers A and D. However, remember that moving in a circle means that there is an acceleration, the centrifugal one, even if the speed does not change. Hence C is also incorrect.
The correct answer is B, since if there is no change in velocity, the frame does not accelerate and it is inertial.
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In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the
yulyashka [42]

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}

4 0
2 years ago
a car moving at 5 m/s accelerates at a rate of 10 m/s2 for 25 seconds. How far does it move during this time?
Alexus [3.1K]

Answer:

t is time in s For example, a car accelerates in 5 s from 25 m/s to 3 5m/s. Its velocity changes by 35 - 25 = 10 m/s. Therefore its acceleration is 10 ÷ 5 = 2 m/s2

Explanation:

3 0
2 years ago
A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.
Simora [160]

Answer:

\mu_k=\frac{a}{g}

Explanation:

The force of kinetic friction on the block is defined as:

F_k=\mu_kN

Where \mu_k is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

N=mg\\F_k=F=ma

Replacing this in the first equation and solving for \mu_k:

ma=\mu_k(mg)\\\mu_k=\frac{a}{g}

6 0
3 years ago
A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 4700 MJ with respect to Eart
erastovalidia [21]

during satellite motion we know that total energy is always conserved

so here we will have

KE_i + PE_i = KE_f + PE_f

here we know that

KE_i = 4800 MJ

PE_i = 4700 MJ

now at other position

PE_f = 6000 MJ

now from above equation we have

4800 +4700 = 6000 + KE

now we have

9500 = 6000 + KE

KE = 9500 - 6000 = 3500 MJ

so its kinetic energy will be 3500 MJ

6 0
3 years ago
A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h
Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

              P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂

             

The pressure when the water comes out is the atmospheric pressure

           P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

           y₂-y₁ = 15 m

Now let's use the continuity equation

             v₁ A₁ = v₂ A₁

The area of ​​a circle is

             A = π r² = π (d/2)²

            v₁ π d₁²/ 4 = v₂ π d₂²/ 4

            v₂ = v₁ d₁² / d₂²

Let's replace

           P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

           P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

           v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

           d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

           d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

            v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2  -9.8 15/2

            v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

            v₁ = √ 2.2477 10⁸ /0.8710

             v₁ = 1,606 10⁴ m / s

4 0
2 years ago
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