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loris [4]
3 years ago
10

Someone Help me please

Physics
1 answer:
natali 33 [55]3 years ago
7 0

Answer:

a) Yes because spiderman isn’t part of the system (the train), so he exerts an external force on the system, causing it to slow. The sign is negative because the train is slowing negative.

b) The train was initially moving, so it had kinetic energy, the energy of movement. There isn’t the same amount anymore because spiderman did work onto the train. Work is defined by the change of kinetic energy of the system, so the train didn’t have the same amount of that type of energy at the end.

I hope this helps! :)

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An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
What wavelength of light (in nm) is associated with a frequency of 8.01E15 Hz?
AlekseyPX

Answer: 37.5 nm

Explanation: speed of light c= 3.00·10^8 m/s.

I use same accuracy to speed of light as it's for frequency.

Frequency f= 8.01·10^15 1/s

Speed c = wavelength · frequency

Wavelength = c/f = 3.745·10^-8 m

6 0
3 years ago
A scientist is studying a shock wave from an earthquake. What kind of wave is being studying?
Pavel [41]

Answer:

Longitudinal Mechanical Wave

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Shock wave, strong pressure wave in any elastic medium such as air, water, or a solid substance, produced by supersonic aircraft, explosions, lightning, or other phenomena that create violent changes in pressure.

Shock waves travel faster than sound and their speed increases as the amplitude of the wave is increased but their intensity fades faster due to the fact that some of its energy gets expended in the form of heat due to the resistance of the medium.

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Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.

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