What occurs as a ray of light passes from<br> al inilo water?

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is $v = 3.79 *10^{5} \ m/s$

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 144 \ kV /m = 144*10^{3} \ V/m$

The magnetic field is $B = 0.38 \ T$

The force due to the electric field is mathematically represented as

$F_e = E * q$

and

The force due to the magnetic field is mathematically represented as

$F_b = q * v * B * sin(\theta )$

Now given that it is perpendicular , $\theta = 90$

=> $F_b = q * v * B * sin(90)$

=> $F_b = q * v * B$

Now given that it is not deflected it means that

$F_ e = F_b$

=> $q * E = q * v * B$

=> $v = \frac{E}{B }$

substituting values

$v = \frac{ 144 *10^{3}}{0.38 }$

$v = 3.79 *10^{5} \ m/s$

7 0

3 years ago

Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P of the gas is inversely pro

jolli1 [7]

Answer:

a) $V=\dfrac{5.3}{P}$

b) $ML^{-4}T^{-2}$ .

Explanation:

Given that

Boyle's law

P V = Constant ,at constant temperature

a)

Given that

$P_1=50KPa$

$V_1=0.106m^3$

We know that for PV=C

$P_1V_1=P_2V_2=PV$

Now by putting the values

PV= 50 x 0.106

$V=\dfrac{5.3}{P}$

Where P is in KPa and V is in $m^3$

b)

PV= C

Take ln both sides

So $\ln(PV)=\ln C$

lnP + lnV =lnC ( C is constant)

By differentiating

$\dfrac{dP}{P}+\dfrac{dV}{V}=0$

So

$\dfrac{dP}{dV}=-\dfrac{P}{V}$

When P= 50 KPa

$\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}$

It indicates the slope of PV=C curve.

It unit is $\dfrac{Pa}{m^3}$ .

Or we can say that $ML^{-4}T^{-2}$ .

5 0

3 years ago

Read 2 more answers

A circuit has a resistance of 10 Ω and current of 5 A. What is the voltage?<br><br>

raketka [301]

Answer: 50 V

Explanation:

$V=IR\\V=(5)(10)\\V=50$

6 0

3 years ago

A football is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of

Andrei [34K]

Solution:

According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.

So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.

v_f=v_o + at ……..(a)

[where v_f and v_o are final velocity and initial velocity, respectively]

Now ,

Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.

Applying this value in equation (a)

v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction

For calculating the magnitude of the equation we have to square root the given value

(16.6i - 17.165y)

\\ \left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ = \sqrt{275.56+294.637225}\\= \sqrt{570.197225}\\= 23.87[/tex]

5 0

3 years ago

Read 2 more answers

PLEASE HELP ME WITH THIS ONE QUESTION

adoni [48]

Answer:

B) R1 = 6 V and R2 = 6V

Explanation:

In series, both resistors will carry the same current.

that current will be I = V/R = 12 / (10 + 10) = 0.6 A

The voltage drop across each resistor is V = IR = 0.6(10) = 6 V

6 0

3 years ago

What occurs as a ray of light passes from al inilo water?