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4 years ago

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A proton, moving in a uniform magnetic field, moves in a circle perpendicular to the field lines and takes time T for each circl

e. If the proton's speed tripled, what would now be its time to go around each circle?
A- 3T
B- T
C- T/3
D- 9T
E- T/9

1 answer:

svet-max [94.6K]4 years ago

5 0

Your answer is c sooooooooooo yea

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In the classic horse and cart problem, a horse is attached to a cart that can roll along on a set of wheels. Which of the follow

Marina86 [1]

E is correct because net force in the forward direction is greater

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3 years ago

This problem has been solved!

RideAnS [48]

Answer:

Option (a)

Explanation:

We will discard options that don't fit the situation:

Option b: <em>Incorrect </em>since if the driver "hits the gas" then velocity is augmenting and it's not constant.

Option c and d: <em>Incorrect </em>since the situation doesn't give us any information that could be related directly to the terrain or movement direction.

Option a: Correct. At <em>stage 1</em> we can assume the driver was going at constant speed which means acceleration is constantly zero. At <em>stage 2 </em>we can assume the driver augmented speed linearly, this is, with constant positive acceleration. At <em>stage 3 </em>we can assume the driver slowed the speed linearly, with constant negative acceleration.

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3 years ago

Be sure to answer all parts. Without stratospheric ozone (O3), harmful solar radiation would cause gene alterations. Ozone forms

Ratling [72]

Answer:

for C-Cl bond $\lambda_1=3.66\times 10^{-9}\ m$

for O-O bond $\lambda_2=2.43\times 10^{-9}\ m$

Explanation:

We have the bond enegy for C-Cl, $BE_1=327\ kJ.mol^{-1}= 327000\times 1.66\times 10^{-22}\ J\ per\ molecule$

bond enegy for C-Cl, $BE_2=494\ kJ.mol^{-1}= 494000\times 1.66\times 10^{-22}\ J\ per\ molecule$

<u>Now as we know the energy of electromagnetic waves is given by:</u>

$E=h.\nu$

here E = BE

<u>Then for C-Cl:</u>

$327000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_1$

$\nu_1=8.18733\times 10^{16}\ Hz$

Now wavelength:

$\lambda_1=\frac{c}{\nu_1}$

$\lambda_1=\frac{3\times 10^8}{8.18733\times 10^{16}}$

$\lambda_1=3.66\times 10^{-9}\ m$

<u>For O2:</u>

$494000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_2$

$\nu_2=1.236863\times 10^{17}\ Hz$

Now wavelength:

$\lambda_2=\frac{c}{\nu_2}$

$\lambda_2=\frac{3\times 10^8}{1.236863\times 10^{17}}$

$\lambda_2=2.43\times 10^{-9}\ m$

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3 years ago

Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

oksano4ka [1.4K]

Answer:

Help me please?

Explanation:

Did you get the answer? I believe it’s either C. +q or D. 0

6 0

3 years ago

Read 2 more answers

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts. The sup

Finger [1]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts. The supply storage area of the lunar outpost, where gravity is 1.63 m/s2, can only support 1 x 10^5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost

Answer:

601,220N

Explanation:

Given that:

Gravity at lunar outpost = 1.63m/s²

Acceleration due to gravity on earth = 9.8m/s²

Supported weight = 1 * 10^5 N

Maximum weight of supplies as measured on earth;

(Ratio of the gravities) * weight of supplies

(9.8m/s² / 1.63m/s²) * (1 * 10^5 N)

6.0122 * (1 * 10^5)

6.0122 * 10^5 N

= 601,220 N

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3 years ago