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3 years ago

14

The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le

ast 8 minutes, then how far does the shuttle go

1 answer:

Vinil7 [7]3 years ago

3 0

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

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Two infinite plane sheets with uniform surface charge densities are placed parallel to each other with separation d. in the regi

kupik [55]

Answer:Uniform between the plates

Explanation:

Electric Field due to infinite sheet with uniform charge is given by

$E=\dfrac{\sigma }{\epsilon _0}$

Where $\sigma$ =charge density

$\epsilon _0$ =Space Permittivity

From the formula we can see that Electric field is independent of distance of the particle, so it is uniform between the plates .

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3 years ago

A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

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Where,

T = Tension

$\rho =$ Linear density

Our data are given by

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Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

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Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

$f = \frac{1}{T}$

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$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

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3 years ago

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo

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Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

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Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

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Answer:

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= 0.0054 ÷ 0.000000152881

=<u>35321.58999 N/C</u>

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2 years ago

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3 years ago

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vagabundo [1.1K]

Answer

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3 years ago