rocks move trough the rock cycle through uplift,folding,and faulting cased by movements of the earth´s_______

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

notka56 [123]

You must assume that the mass of the rocket and engine remains constant - even though the engine is burning.

You know the engine produces 13.8N for a distance of 14.6m

The total energy expended (work done) by the engine is FxD so you can calculate that

Now - some of that is given to the rocket as kinetic and potential energy, and some is expended against the drag force.

At the peak of its flight ALL the energy given to the rocket is potential energy (its velocity is zero) and that is calculated as mgh

So
Energy given to rocket = mgh
Energy expended by engine = F x D (D= height where engine stops)
Energy 'lost' to drag is the difference between the two values.

7 0

3 years ago

Read 2 more answers

Write the correct answers:)

Darina [25.2K]

Answer:

a) iv. Displacement per unit time

b) ii. m/s2

c) this question is wrong

d) iv. m

6 0

3 years ago

What are the workings of a electromagnet?<br> plssssssssssssssssssssss answer<br> its U.R.G.E.N.T

Rudiy27

Answer:

Wh en an ele ctric c urrent flows in a wire, it cre at es a ma g ne tic fie ld around the wire. Thi s ef fect can be used to make an electromagnet . A simple electr om agnet a length of wire tur ned into a c oil an d co nnected to a batt e ry or po wer sup ply.

Explanation:

8 0

4 years ago

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.220 T. If the kinetic energy of the electr

Gwar [14]

Answer:

$971605.66\ \text{m/s}$

$25.1\ \mu\text{m}$

Explanation:

m = Mass of electron = $9.11\times 10^{-31}\ \text{kg}$

B = Magnetic field = 0.22 T

K = Kinetic energy of electron = $4.3\times 10^{-19}\ \text{J}$

q = Charge = $1.6\times 10^{-19}\ \text{C}$

v = Velocity of electron

r = Radius of curved path

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4.3\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=971605.66\ \text{m/s}$

The speed of the electron is $971605.66\ \text{m/s}$

The force balance of the system is given by

$qvB=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{mv}{qB}\\\Rightarrow r=\dfrac{9.11\times 10^{-31}\times 971605.66}{1.6\times 10^{-19}\times 0.22}\\\Rightarrow r=0.0000251=25.1\ \mu\text{m}$

The radius of the curved path is $25.1\ \mu\text{m}$

6 0

3 years ago

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

3 years ago

rocks move trough the rock cycle through uplift,folding,and faulting cased by movements of the earth´s________