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rosijanka [135]

3 years ago

13

KATZPSEF1 30.P.041. My Notes Ask Your Teacher A charged particle enters a region of space with a uniform magnetic field B = 1.98

i T. At a particular instant in time, it has velocity v = (1.46 ✕ 106 i + 2.42 ✕ 106 j) m/s. Based on the observed acceleration, you determine that the force acting on the particle at this instant is F = 1.60 k N. What are the following?(a) the sign of the charged particle
(b) the magnitude of the charged particle

1 answer:

sladkih [1.3K]3 years ago

6 0

Answer: The charge on the particle is positive

While the magnitude = 0.00028C

Explanation:

Please find the attached file for the solution

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A motorcycle begins at rest and accelerates uniformly S7.9 we want to find a time to take the motorcycle to reach a speed of 100

Len [333]

The motorbike reaches 100 km/h in 3.5 seconds

Explanation:

The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the motorbike in this problem,

u = 0 (it starts from rest)

$v = 100 km/h = 27.8 m/s$ is the final velocity

$a=7.9 m/s^2$ is the acceleration

Solving for t, we find the time it takes for the bike to reach that velocity:

$t=\frac{v-u}{a}=\frac{27.8-0}{7.9}=3.5 s$

Learn more about accelerated motion:

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6 0

3 years ago

Which of the following does NOT describe the conversion of potential to kinetic energy?

ivann1987 [24]

I believe it would be A. (:

8 0

3 years ago

One of the primary goals of the Kepler space telescope is to search for Earth-like planets. Data gathered by the telescope indic

amid [387]

Answer:

85.62 m

168.75 years

101.04 years

Explanation:

$L_0$ = Length of ship = 143 m

v = Velocity of ship = 0.8c

c = Speed of light

s = Distance to Boralis orbit = 135 ly

Gamma value

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \gamma=\dfrac{1}{\sqrt{1-\dfrac{0.8^2c^2}{c^2}}}\\\Rightarrow \gamma=1.67$

Length contraction is given by

$L=\dfrac{L_0}{\gamma}\\\Rightarrow L=\dfrac{143}{1.67}\\\Rightarrow L=85.62\ m$

The length is 85.62 m

Time taken

$t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{135}{0.8}\\\Rightarrow t=168.75\ years$

Time taken from the perspective one Earth is 168.75 years

Time dilation is given by

$t'=\dfrac{t}{\gamma}\\\Rightarrow t'=\dfrac{168.75}{1.67}\\\Rightarrow t'=101.04\ years$

The time taken from the perspective of the ship is 101.04 years

8 0

3 years ago

Compare and Contrast the two main branches of physical science. How are they the same? How are they different?

nignag [31]

Physics
Chemistry

Physical science is science which studies non-living things.This covers things on earth to the universe.The main branches are physics and chemistry.The other sub-branches are Astronomy, meteorology and Geology.Physics is concerned with the basic principals of matter and energy.It covers concepts in motion, energy, force, pressure , light and many other areas we encounter in our day to day life.Chemistry is concerned with composition of matter and how it reacts.Material in chemistry is understood through observing molecules, atoms, elements and their interaction in nature including their properties.Chemistry is further divided into organic chemistry and natural chemistry sub-branches.

Hope this helps you!!!

8 0

2 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

3 years ago