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Keith_Richards [23]
3 years ago
7

Why is velocity most important in space launch?

Physics
2 answers:
Debora [2.8K]3 years ago
6 0
Escape velocity is the velocity an object needs to escape the gravitational influence of a body if it is in free fall, i.e. no force other than gravity acts on it. Your rocket is not in free fall since it is using its thruster to maintain a constant velocity so the notion of "escape velocity" does not apply to it.
mote1985 [20]3 years ago
6 0

Answer: Escape Velocity

Explanation:

Escape velocity is the velocity an object needs to escape the gravitational influence of a body if it is in free fall, i.e. no force other than gravity acts on it. Your rocket is not in free fall since it is using its thruster to maintain a constant velocity so the notion of "escape velocity" does not apply to it.

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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
Young's Modulus refers to changes in the a Volume b- Length c- Body layers
Novosadov [1.4K]

Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

Normal stress is given by force per unit area. Longitudinal strain is the change in length per unit original length.

The mathematical definition of Young's modulus is given by :

Y=\dfrac{F/A}{\Delta L/L}..........(1)

Where

\Delta L is the change in length

F is the force

A is the area of cross section

So, the Young's modulus refers to the change in length of the object. Hence, the correct option is (b) "length".

8 0
3 years ago
The wave length of violet light rounded to the nearest nanometer is a __ nm
nataly862011 [7]
Wavelength= speed / frequency 

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7) 

in nanometer = 413 nm
6 0
3 years ago
Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o
muminat

As per above given data

initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

\Delta v = v_f - v_i

\Delta v = -18.8 - 19.3

\Delta v = -38.1 km/s

\Delta v = -3.81 * 10^4 m/s

Part b)

Now we need to find acceleration

acceleration is given by formula

a = \frac{\Delta v}{\Delta t}

given that

\Delta v =- 3.81 * 10^4 m/s

\Delta t = 2.07 years = 6.53 * 10^7 s

now the acceleration is given as

a = \frac{-3.81 * 10^4}{6.53 * 10^7}

a = - 5.84 * 10^{-4}m/s^2

so above is the acceleration

4 0
3 years ago
A heavy rope, 80 ft long and weighing 32 lbs, hangs over the edge of a building 100 ft high. how much work w is done in pulling
allochka39001 [22]
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
 We have then:
 W = Fd
 The distance varies, so we must integrate:
 from 0 to 20:
 W = ∫F (x) dx
 W = ∫32xdx
 W = 32∫xdx
 W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
 answer:
 6400 ft * lbs is work done pulling the rope up 20 ft
6 0
4 years ago
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