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joja [24]
4 years ago
15

A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.30 m/s2. At 20.0 s after bla

stoff, the engines suddenly fail, and the rocket begins free fall. which means that the force they produce instantly stops.(a) How high above the launch pad will the rocket eventually go?(b) Find the rocket's velocity at its highest point.(c) Find the magnitude of the rocket's acceleration at its highest point.
Physics
1 answer:
lesya [120]4 years ago
6 0

Answer:

Explanation:

v = u +at

u = 0

a = 2.3 m /s²

t = 20 s

v = 2.3 x 20

= 46 m /s

Distance covered under acceleration of 2.3 m/s²

s = ut + 1/2 at²

= 0 + .5 x 2.3 x 20²

= 460 m

After that it moves under free fall ie g acts on it downwards .

v² = u² - 2gh , h is height moved by it under free fall

0 = 46² - 2 x 9.8 h

h = 107.96 m

Total height attained

= 460 + 107.96

= 567.96 m

b ) At its highest point ,it stops so  its velocity = 0

c ) rocket's acceleration at its highest point = g = 9.8 downwards .

At highest  point , it is undergoing free fall so its acceleration  = g

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Strong electrical currents in close proximity to the magnet.

Other magnets in close proximity to the magnet.

Neo magnets will corrode in high humidity environments unless they have a protective coating.

Explanation: Heat radiation

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3 years ago
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A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
Challenge: Determine the speed of the waves at each tension setting (high, medium and low). Explain what measurements you made t
Svetradugi [14.3K]

Answer:

high tension: 4.2 × 1.5 = 6.3 cm/s

medium tension: 2.8 ×1.5 = 4.2 cm/s

low tension: 0.8 × 1.5 = 1.2 cm/s

Explanation: Given Settings:

amplitude: 0.75 cm

damping: zero

Using

Speed = frequency ×wavelength

Where

Wavelength = 0.75 × 2 = 1.5 cm

Therefore:

high tension: 4.2 × 1.5 = 6.3 cm/s

medium tension: 2.8 ×1.5 = 4.2 cm/s

low tension: 0.8 × 1.5 = 1.2 cm/s

4 0
3 years ago
Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of (20.0m/s)ı^−(4.0m/s)ȷ^(20.0m/s)ı^−(4.0m/s
Art [367]

Answer:

(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^

Explanation:

Given

W = 0.56N = mg

m = 0.56/g = 0.56/9.8 = 0.057kg

t = 3.00ms = 3.00×10-³s

Impulse is a vector quantity so we would treat it as such

We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that

u = (20.0m/s)ı^−(4.0m/s)ȷ^

ux = 20m/s

uy = – 4.0m/s

F = – (380N)ı^+(110N)ȷ^

Fx = –380N

Fy = 110N

J = impulse = force × time = F×t

So Jx = Fx ×t

Jy = Fy×t

Jx = –380×3×10-³ = -1.14Ns

Jy = 110×3×10-³ = 0.330Ns

Impulse also equals the change in momentum of the body. So

J = m(v–u)

J/m = v – u

V= J/m + u

Vx = Jx/m + ux

Vx = –1.14/0.057 + 20

Vx = -20 + 20 = 0m/s

Vx = 0m/s

Vy= Jy/m + uy

Vy= 0.33/0.057 + (-4.0)

Vy= 5.79 + (-4.0) = 1.79m/s

V = (0m/s)ı^−(1.79m/s)ȷ^

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3 years ago
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Marrrta [24]

Answer:

Explanation:

Coordinate system is one that describe the location of an object in a given plane. It implies the use of axes (coordinates) and points.

Given that the man in the question walks 400 m due north of east. The cardinal points can be used in this case, with the north and east cardinals as the required axis.

scale = \frac{length on drawing}{original length}

         = \frac{10}{400}

         = \frac{1}{40}

scale = 1:40

This is a reduced scale which implies that 1 cm on the drawing is equal to 40 m on the original length.

The man's direction is 45^{o} north of east.

The graphical drawing of the vector is herewith attached to this answer.

5 0
3 years ago
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