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salantis [7]
3 years ago
13

A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.

The bottom of the spring, whose spring constant is 480 N/mN/m , is anchored to the ground. Part A Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed

Physics
1 answer:
blondinia [14]3 years ago
7 0

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N

According to Hooks Law

F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m

The part b and c of this question is done in the attachment

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
3 years ago
What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?
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Answer:

0.044 V

Explanation:

E = Electric field = 5.5\times 10^6\ V/m

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V

The voltage across the membrane is 0.044 V

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To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.
Karolina [17]

Answer:

a)  T² = (\frac{4\pi ^2}{GM})  r³

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kinetic energy  depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

            F = ma

in this case the acceleration is centripetal

            a = v² / r

the linear and angular variable are related

           v = w r

we substitute

           a = w² r

force is the universal force of attraction

           F = G \frac{m M}{r^2}

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         G \frac{m M}{r^2} = m w^2 r

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         w = 2π f = 2π / T

we substitute

            ( \frac{2\pi }{T} ) = \frac{GM}{r^3}

the final equation is

             T² = ()  r³

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           v = w r

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            v = \sqrt{\frac{GM}{r} }

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Kinetic energy

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           K = ½ M GM / r

           K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

          U =

          U = -G mM / r

dependency is the inverse of distance

Angular momentum

          L = r x p

for a circular orbit

           L = r p = r Mv

           L =

         L =

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3 years ago
Which astronomer gave his name to a telescope launched into space in 1990?
Mnenie [13.5K]

Answer:

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Explanation:

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3 years ago
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