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Gala2k [10]
3 years ago
6

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward

acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s. How long did the burn phase last?
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
3 0
The important thing to note here is the direction of motion of the test rocket. Since it mentions that the rocket travels vertically upwards, then this motion can be applied to rectilinear equations that are derived from Newton's Laws of Motions.These useful equations are:

y = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
y is the vertical distance travelled
v₁ is the initial velocity
v₂ is the final velocity
t is the time 
a is the acceleration

When a test rocket is launched, there is an initial velocity in order to launch it to the sky. However, it would gradually reach terminal velocity in the solar system. At this point, the final velocity is equal to 0. So, v₂ = 0. Let's solve the second equation first.

a = (v₂-v₁)/t
a = (0-30)/t
a = -30/t

Let's substitute a to the first equation:
y = v₁t + 1/2 at²
49 = 30t + 1/2 (-30/t)t²
49 = 30t -15t
49 = 15 t
t = 49/15
t = 3.27 seconds

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Answer:

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ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

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Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

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A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

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Read 2 more answers
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnet
Mumz [18]
<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb\sqrt{b^2 + L^2})                -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{25.0004})

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m  on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

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3 years ago
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