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2 years ago

12

In projectile mtion, what is the x-component of the initial velocity? if V= Vi = 100 m/s and the angle with horizontal axis Θ =

60 degrees

2 answers:

AleksandrR [38]2 years ago

3 0

Answer:

???????????????????????

torisob [31]2 years ago

3 0

Explanation:

The x - component of the initial velocity is given as:

Ux = Vi cos theta

Ux = 100cos 60

Ux = 50m/s

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The noise level coming from a pig pen with

sweet [91]

Answer:

The decibel of the remaining pigs is 51.5 dB.

Explanation:

Decibel (dB) is a unit of measure of the intensity of a given sound.

Number of pigs = 199, noise level = 74.3 dB.

Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:

I $\alpha$ N

I = kN

where k is the constant of proportionality.

⇒ k = $\frac{I}{N}$

= $\frac{74.3}{199}$

k = 0.3734

When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.

Thus, the becibel level (I) of the remaining pigs can be determined by:

I = kN

= 0.3734 × 138

= 51.53 dB

The becibel level (I) of the remaining pigs is 51.53 dB.

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3 years ago

What is always true of a liquid's shape and volume?

Natalija [7]

Molecules of liquid are far away from each other but less compared to gas. Liquid has the ability to flow. Liquid molecules can move. Liquid occupies the shape of the container. Thus, they do not have fixed shape. Liquid has fixed volume because they are not compressible(easily)

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3 years ago

A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi

motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

3 0

2 years ago

A batter hits a ball and it is caught 4 seconds later 100 m from home plate. What is the initial velocity (vector!!!) of the bal

Eddi Din [679]

Answer:

The initial velocity vector of the ball is;

$\underset{u}{\rightarrow}$ = 25·i + 19.6·j

Explanation:

The given parameters are;

The time of flight of the ball = 4 seconds

The horizontal distance from the plate at which the ball is caught = 100 m

Let 'u', represent the initial velocity, we have;

u × cos(θ) × t = u × cos(θ) × 4 s = 100 m

u × cos(θ) = 25 m/s...(1)

$t = \dfrac{2 \cdot u \cdot sin(\theta)}{g} = \dfrac{2 \cdot u \cdot sin(\theta)}{9.8 \ m/s^2} = 4 \ seconds$

∴ 2·u·sin(θ) = 9.8 m/s² × 4 s = 39.2 m/s

$\therefore \dfrac{2 \cdot u \cdot sin(\theta)}{u \cdot cos(\theta)} = 2 \cdot tan(\theta) = \dfrac{39.2 \ m/s}{25 \ m/s} = 1.568$

tan(θ) = 1.568/2 = 0.784

θ = arctan(0.784) ≈ 38.096°

The direction of the velocity vector of the ball, θ ≈ 38.096°

From equation (1), we have;

u × cos(θ) = 25 m/s

∴ u = 25 m/s/cos(θ) = 25 m/s/cos(arctan(0.784)) = 31.7672787629 m/s

The magnitude of the initial velocity vector, u = 31.7672787629 m/s

The vertical component of the initial velocity, $u_y$ = u × sin(θ)

∴ $u_y$ = 31.7672787629 × sin(arctan(0.784))

Therefore, the initial velocity vector of the ball is approximately, v = 31.767 m/s in a direction 38.096° above the horizontal, from which we have;

u = uₓ·i + $u_y$ ·j = 25·i + 19.6·j

$\underset{u}{\rightarrow}$ = 25·i + 19.6·j

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2 years ago

The asthenosphere is a layer whose distinctiveness from the rest of the mantle is based on its ___________.

gavmur [86]

<span>reduced rigidity should be the answer

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8 0

3 years ago

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