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aev [14]
3 years ago
10

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Give your answer in centimetres.

Physics
1 answer:
Juliette [100K]3 years ago
8 0

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

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Answer:

I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

θ = (225 - 180) ± arcsin(13/30)

θ = 45 ± 25.679...

70.679 ≤ θ ≤ 19.321

components of vector B would be

Bx = |B|cosθ

By = |B|sinθ

3 0
2 years ago
A tennis racket strikes a tennis ball with a force of 100 N. With what force
viva [34]

Answer:

-100N

Explanation:

Newton's third law of motion states that to every force exerted on one body, there is an equal and opposite force. This means that if object A exerts an ACTION force on B, there is a force called REACTION FORCE, which is equal and opposite, exerted on A by B.

The action and reaction forces are equal in size/magnitude but opposite in direction. In this case where a tennis racket strikes a tennis ball with a force (action force) of 100N, the ball will strike the racket with a reaction force of -100N.

F(RB) = -F(BR)

8 0
3 years ago
what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s
yaroslaw [1]

Answer : The value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = ?

t = time = 17s

[A_t] = final concentration = 0.0981 M

[A_o] = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}

k=0.51M^{-1}s^{-1}

Therefore, the value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

6 0
3 years ago
What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?
anygoal [31]
D. 35n forwards....................
8 0
3 years ago
a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

Heat Supplied Q=10 J

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

thus \Delta U=mc_v\Delta T=Q

c_v for mono-atomic gas is 12.471 J/K-mol

n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

3 0
3 years ago
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