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astraxan [27]
3 years ago
14

A hobby rocket reaches a height of 72.3 meters and lands 111 meters from the launch point

Physics
1 answer:
faust18 [17]3 years ago
8 0
Using the following formulas for projectile motion:

Height, H = ( Vo^2 * sin theta^2 )/g

Range, R = ( Vo^2 * sin 2*theta )/g

Rearranging in terms of Vo^2:

Vo^2 = gH / sin theta^2

Vo^2 = gR / sin 2*theta

Equating the two formulas to each other to solve for the angle theta:

gR / sin 2*theta = <span>gH / sin theta^2
</span>
Substituting the given values:

(9.8)(111) / sin 2*theta = (9.8)(72.3)<span> / sin theta^2
</span>angle = 52.36 degrees

Therefore, the angle of launch is approximately 52.36 degrees.
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A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st
OLEGan [10]

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

8 0
3 years ago
g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo
mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D)  Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 ,  exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0
2 years ago
The symbol used for magnetic field is:
alina1380 [7]
The letter "B" is used for magnetic field
5 0
3 years ago
Select the correct answer what is meditation
german

Answer:

Meditation can be defined as a set of techniques that are intended to encourage a heightened state of awareness and focused attention. Meditation is also a consciousness-changing technique that has been shown to have a wide number of benefits on psychological well-being.

3 0
2 years ago
When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil
Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

2t = 2\lambda_{film}

t = \lambda_{film}

The index of refraction of soap is given, then

\lambda_{film} = \frac{\lambda_{vacuum}}{n}

Combining the results of all steps we get

t = \frac{\lambda_{vacuum}}{n}

Rearranging, we find

\lambda_{vacuum} = tn

\lambda_{vacuum} = (278)(1.33)

\lambda_{vacuum} = 369.74nm

4 0
2 years ago
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