Ask question

Ask question

All categories

3 years ago

14

A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

s velocity. What is the magnitude of the force that the proton experiences while it moves through the magnetic field?

1 answer:

scZoUnD [109]3 years ago

3 0

A charged particle moving in a magnetic field experiences a force equal to:

$\vec{F}=q\vec{v}\times \vec{B}$

Thus, the magnitude of the force that the proton experiences is given by:

$F=qvBsin\theta$

The magnetic field is perpendicular to the proton's velocity, therefore, we have $\theta=90^\circ$ . Replacing the given values, we obtain:

$F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N$

You might be interested in

A net force of -1,000 Newtons is delivered to the object over a time of .02 seconds. Calculate the new velocity of the object.

MariettaO [177]

The change in velocity (v₂ - v₁) is

<em> (-20) / (the object's mass)</em>.

Call it a crazy hunch, but I can't shake the feeling that there was more
to the question before the part you copied, that mentioned the object's
mass, and its velocity before this force came along.

4 0

3 years ago

What are the characteristics of high energy wave?

Ivahew [28]

Answer:

D. High frequency and short wavelengths.

Explanation:

If a wave is high in energy it will have a higher frequency.

High frequency = short wavelengths

8 0

2 years ago

Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t

ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body Mass (gram) Engine Force (newtons)

1 500 1 25

2 1500 2 20

3 750 3 30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

20 = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

$=\frac{200}{5}$

$= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $40 \ m/s^2$ at the start of the launch.

8 0

3 years ago

An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,

insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

$Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}$

Where

$t_{hf}$ = Temperature at the inside of the pipe = 300 °C

$t_{f}$ = Temperature at the outside of the pipe = 20 °C

r₁ =internal radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

$k_A$ = 15 W/m·K

$k_B$ = 0.038 W/m·K

$h_{hf}$ = 75 W/m²·K

$h_{cf}$ = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

$Q = \frac{t_A-t_B}{R_T}$ Where $R_T$ for the air film on the pipe outer surface is given by

$R_T= \frac{1}{\alpha A}$

where A =area of the outside of the pipe

= $\frac{1}{10*2\pi*0.07*1 }$ = 0.227 K/W

Therefore

131.32 W = $\frac{t_A-20}{0.227}$ which gives

$t_A$ = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0

3 years ago

The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th

ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, $B=2\ mT=2.2\times 10^{-3}\ T$

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :

$B=\dfrac{\mu_oNI}{2\pi r}$

$r=\dfrac{\mu_oNI}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}$

r = 0.00523 m

or

$r=5.23\times 10^{-3}\ m$

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers