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3 years ago

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A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

0.00687 T 0.00687 T and the proton's velocity makes an angle of 101 ∘ 101∘ with the field, what is the magnitude of the magnetic force acting on the proton?

1 answer:

iogann1982 [59]3 years ago

4 0

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

c = Speed of light = $3\times 10^8\ m/s$

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$

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A solenoid of length 2.50 cm and radius 0.750 cm has 25 turns. If the wire of the solenoid has 1.85 amps of current, what is the

vfiekz [6]

Answer:

13.875 T

Explanation:

Parameters given:

Length of solenoid, L = 2.5 cm = 0.025 m

Radius of solenoid, r = 0.75 cm = 0.0075 m

Number of turns, N = 25 turns

Current, I = 1.85 A

Magnetic field, B, is given as:

B = (N*r*I) /L

B = (25 * 0.0075 * 1.85)/0.025

B = 13.875 T

4 0

3 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago

Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each

qaws [65]

Answer:

A large piece of charcoal on a grill in the sunlight (if it's burning) will consist of the following types of energy:

Chemical
Heat and
Light

Explanation:

Charcoal is basically carbon which is produced when wood is heated strongly in the absence of oxygen. From a chemistry point of view, charcoal contains combustible carbon whose chemical formula is C. Sometimes, $H_{2}O$ which is water may be found in it but in very small units.

All matter contains Heat energy. Charcoal is not an exception. As the charcoal burns, the heat energy is produced along with Light energy.

Light comes in many forms such as Infrared rays, Xrays, Visible Spectrum light, etc.

The glow which the coal gives off fall under the visible spectrum of light.

Cheers

8 0

2 years ago

Write two units for electric field intensity and show thier equivalence?

nataly862011 [7]

Answer:

Electric field intensity is the force experienced by a test charge q in a electric field E.

7 0

3 years ago

Read 2 more answers

Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—th

krek1111 [17]

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

7 0

3 years ago