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Sergeeva-Olga [200]

3 years ago

14

A charge enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc

of radius R. If the charged particle enters the same magnetic field with three times the speed, what will be the radius of the circular arc?

1 answer:

frozen [14]3 years ago

7 0

Answer:

The new radius is three times the initial radius

Explanation:

The magnetic force is given by the equation

F = q v x B

Where the blacks indicate vectors, q is the electric charge, v the velocity of the particle and B the magnetic field. The scalar magnitude of this force is

F = q v B sin θ

Where θ is the angle between the velocity and the magnetic field.

As the force is perpendicular to the velocity, the particle describes a circular motion, using Newton's second law we can find the acceleration that is centripetal.

F = ma

a = v² / r

q v B = mv² / r

R = mv / qB

Let's calculate for the new speed (v₂ = 3v)

R₂ = (m / qB) 3v

R₂ = 3 R

R₂ / R = 3

The new radius is three times the initial radius

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A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.

Jobisdone [24]

Answer:

<em>(a) t = 4.52 sec</em>

<em>(b) X = 1,156.49 m</em>

Explanation:

<u>Horizontal Launching </u>

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity $v_o$ remains the same in time. The distance x is computed as .

$\displaystyle x=v_o.t$

(a)

The vertical component of the velocity $v_y$ starts from zero and gradually starts to increase due to the acceleration of gravity as follows

$v_y=gt$

This means the vertical height is computed by

$\displaystyle h=h_o-\frac{gt^2}{2}$

Where $h_o$ is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0

$\displaystyle t=\sqrt{\frac{2h_o}{g}}$

$\displaystyle t=\sqrt{\frac{2(100)}{9.8}}=4.52\ sec$

(b)

We now compute the horizontal distance knowing $v_o=256\ m/s$

$\displaystyle x=(256).(4.52)$

$X=1,156.49\ m$

4 0

3 years ago

A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?

ser-zykov [4K]

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

${ \rm{ {v}^{2} = {u}^{2} + 2gs}}$

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

${ \rm{0 {}^{2} = {2}^{2} - (2 \times 9.8 \times s)}} \\ \\ { \rm{4 = 19.6s}} \\ \\ { \rm{s = 0.204 \: m}}$

4 0

2 years ago

Read 2 more answers

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

Is visible light a form of<br> radiation? Explain.

natali 33 [55]

Answer:

Yes,in fact visible 'light' is a form of radiation, which can be defined as an energy that travels in the form of electromagnetic waves. It can also be described as a flow of particle-like 'wave-packets', called photons, that travel constantly at the speed of light (about 300 000 kilometres per second).

Explanation:

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3 years ago

an airplane is traveling at an altitude of 31,360. a box of supplies is driped from the cargo hold how long will it take to reac

VMariaS [17]

Answer: 1,600 seconds

Explanation:

31,360/9.8 = 3,200.

Then divide 3,200/2 = 1,600

7 0

3 years ago