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Blizzard [7]
3 years ago
10

An electron is projected vertically upward with a speed of 1.70 ✕ 106 m/s into a uniform magnetic field of 0.320 t that is direc

ted horizontally away from the observer. describe the electron's path in this field when viewed from above.
Physics
1 answer:
Serggg [28]3 years ago
7 0
The electron's path in the magnetic field is a straight line when viewed from above.

In fact, the electron initially moves upward, while the magnetic field is directed horizontally. The electron experiences a force due to the magnetic field (the Lorentz force), whose direction is given by the right-hand rule:
- index finger --> initial direction of the electron (upward)
- middle finger --> direction of the magnetic field (horizontally, away from the observer)
- opposite direction to the thumb* --> direction of the force (horizontally, but perpendicular to the magnetic field, to the right)

This means that the Lorentz force makes the electron moving perpendicular to the magnetic field in the horizontal plane, and since the direction of the field is not changing, this force does not change its direction, so the electron moves in the same direction of the force in the horizontal plane (to the right), therefore following a straight line.

* the direction should be reversed because the charge is negative.
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One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
2 years ago
A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho
amid [387]

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

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Answer:

m=41 kg

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R=2,1 m

F-?

a=v²/R

a=(0,02) ²/2,1=0,0002

F=m*a

F=41*0,0002=0,0082 H

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