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3 years ago

14

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60

-T magnetic field. What is the radius of the resulting path

2 answers:

marysya [2.9K]3 years ago

6 0

Answer:

0.012m

Explanation:

To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:

$r=\frac{mv}{qB}$

m: mass of the proton 9.1*10^{-31}kg

q: charge of the proton 1.6*10^{-19}C

B: magnitude of the magnetic field 0.60T

v: velocity of the proton

In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:

$qV=E_k=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.5*10^{3}V)}{1.67*10^{-27}kg}}=6.92*10^{5}\frac{m}{s}$

Then, by replacing in the formula for the radius you obtain:

$r=\frac{(1.67*10^{-27}kg)(6.92*10^5\frac{m}{s})}{(1.6*10^{-19}C)(0.60T)}=0.012m$

hence, the radius is 0.012m

Bumek [7]3 years ago

5 0

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

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At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/

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Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

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V1 = 6.667 •i m/s

4 0

3 years ago

Now Abel and Kato use what they learned to answer the following problem. The initial speed of a tennis ball is 54 m/s and the la

34kurt

Answer:

h=19.4m

R=199.07 m

Explanation:

To solve this problem we use the parabolic motion equations:

We define:

$v_{i}$ total initial speed =54 $\frac{m}{s}$

$\alpha$ =angle that forms the total initial speed with the horizontal line= 21°

$v_{ix}$ :initial speed component in horizontal direction

= $v_{i} cos\alpha$ =54*cos 21= 50.41 m/s

$v_{iy}$ :initial speed component in vertical direction

= $v_{i} sin\alpha$ =54*sin21=19.35 m/s

$v_{x}$ :horizontal speed at any point on the parabolic path

$v_{y}$ : vertical speed at any point on the parabolic path

g= acceleration of gravity= 9,8 $\frac{m}{s^{2} }$

Equation of the speed of the football in the vertical direction :

$(v_{y} )^{2} =(v_{yi} )^{2} -2*g*y$ Equation (1)

Calculation of the maximum height(h)

The speed of the ball (vy) in the vertical direction gradually decreases until its value is zero when it reaches the maximum height.

We replace y=h, $v_{y} =0$ , $v_{iy} = 19.39\frac{m}{s}$ , $g=9.8\frac{m}{s^{2} }$ in the equation(1)

$0=19.35^{2} -2*9.8*h$

$h=\frac{19.35^{2} }{2*9.8}$

h=19.1 m

Calculating of the range (R)

Fórmula: $R=v_{ix} *t (m)$ Equation (2)

R is the maximum horizontal distance the ball reaches.

The time (t) for the ball to reach R is twice the time the ball spends to reach the maximum height. Then, we calculate the time $(t_{h} )$ ) when the ball reaches the maximum height

We apply the following equation to calculate $t_{h}$ :

$v_{y} =v_{iy} -g*t_{h}$

$0=19.35-9.8*t_{h}$

$9.8*t_{h} =19.35$

$t_{h} =\frac{19.35}{9.8}$

$t_{h} =1.97 s$

$t=2*t_{h}$

$t=3.95 s$

We replace $v_{ix} =50.41 \frac{m}{s}$ and $t=3.95\frac{m}{s}$ in the equation (2)

R=50.41*3.95

R=199.07 m

6 0

3 years ago

A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?

Semenov [28]

Answer:

The maximum range $R_{max}= 132. 72$ m

Explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

$R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }$ m

Where,

g - acceleration due to gravity

Substituting the values in the above equation

$R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }$

= 132.72 m

Hence, the car lands at a distance, $R_{max}= 132. 72$ m

3 0

4 years ago