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Masteriza [31]
3 years ago
9

A plane flying against a jet stream will travel faster than a plane traveling with a jet stream. Please select the best answer f

rom the choices provided T F
Physics
2 answers:
Yuliya22 [10]3 years ago
7 0

The simplified answer is, It is False

stepan [7]3 years ago
3 0

"with the wind" is a tail-wind, and the speeds are added to get the groundspeed.

"against the wind" is a head-wind, and the windspeed is subtracted from the airspeed.

You might be interested in
Please help me thank you !!!!
Fantom [35]

Answer:

<em>UP</em>

Explanation:

heat flows from higher level to lower level

( higher concentration to lower concentration )

and since temperature in above block is less than the lower block, the heat will flow from lower block to higher block .

( Up )

5 0
3 years ago
Which step of the scientific methood do you perform after you state the problem
marshall27 [118]

After stating the problem, we form a hypothesis, or an explanation that can be tested.

8 0
3 years ago
A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan
Nesterboy [21]

Answer:

Compression distance: d \approx 0.102\,m

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (K), in joules, at the expense of elastic potential energy (U), in joules, and overcoming work losses due to friction (W_{l}), in joules:

K + W_{l} = U (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2} (2)

Where:

m - Mass of the block, in kilograms.

v - Final velocity of the block, in meters per second.

\mu - KInetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s - Width of the rough patch, in meters.

k - Spring constant, in newtons per meter.

d - Compression distance, in meters.

If we know that m = 1.2\,kg, v = 2.3\,\frac{m}{s}, \mu = 0.44, g = 9.807\,\frac{m}{s^{2}}, s = 0.05\,m and k = 730\,\frac{N}{m}, then the compression distance of the spring is:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2}

m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}

d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }

d \approx 0.102\,m

4 0
3 years ago
If the acceleration of a truck over a given time interval is zero, how does the instantaneous velocity of that truck at any inst
klemol [59]

Answer:

B

Explanation:

8 0
4 years ago
Determine the frequency of light whose wavelength is 4.257 x 10-7 cm
marin [14]

v =  f (wavelength)i don't know what symbol ya'll use for wavelength so i just put the word instead.We use the greek symbol lambda.So just plug in everything you know.
wavelength=4.257×10^-7x10^-2 and
v=speed of light = 3×10^8
So you should get f= 7.04 ×10^15Hz
7 0
3 years ago
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