Density is determined by:

A proton enters a magnetic field (B = 5.00 x 10^-2 T INTO THE PAGE) with

poizon [28]

Answer:

r = 1.16 m

Explanation:

It is given that,

Magnetic field, $B=5.6\times 10^6\ m/s$

Velocity of proton, $v=5.6\times 10^6\ m/s$

We need to find the radius (in m) of the circular path the particle will follow when it is travelling within the magnetic field. The radius followed by the proton is given by :

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

m is mass of proton, $m=1.67\times 10^{-27}\ kg$

So,

$r=\dfrac{1.67\times 10^{-27}\times 5.6\times 10^6}{1.6\times 10^{-19}\times 5\times 10^{-2}}\\\\r=1.16\ m$

So, the radius of circular path is 1.16 m.

5 0

3 years ago

Suppose an ideal gas is in a container 8.0 l in volume and allowed to expand by having a piston move. The final volume after the

wariber [46]

the initial condition is given as -

$p_{1}$ =33,000 pa

$v_{1}$ =8.0 litre i.e 8× $10^{-3}$ $m^{3}$

$t_{1}$ =300k

final state is given as-

$v_{2} =16 litre$ i.e16 $*10^{-3}$ $m^{3}$

$p_{2} =?$

$t_{2}$ =300k[the temperature is constant here]

here the temperature is constant.we have to calculate the final pressure

as per the Boyle's law,

$p_{1} *v_{1} =p_{2} *v_{2}$

$p_{2} =\frac{p_{1}*v_{1} }{v_{2} }$

= $\frac{8*10^{-3}*33000 }{16*10^{-3} }$

=16500 pa [ans]

7 0

3 years ago

A particle leaves the origin with a speed of 2.1 times 106 m/s at 30 degrees to the positive x axis. It moves in a uniform elect

Verizon [17]

Answer:

-1449.69404 N/C

Explanation:

u = Velocity of particle = $2.1\times 10^6\ m/s$

$\theta$ = Angle = 30°

x = Distance = 1.5 cm

m = Mass of electron = $9.11\times 10^{-31}\ kg$

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

In the case of projectile motion

$x=utcosA\\\Rightarrow t=\dfrac{x}{ucosA}$

The force of on the particle will balance the Electric force

$ma=qE\\\Rightarrow a=\dfrac{qE}{m}$

Now

$y=utsin\theta-\dfrac{1}{2}at^2\\\Rightarrow y=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2$

If y = 0

$0=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow utsin\theta=\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow t=\dfrac{2musin\theta}{qE}$

$\dfrac{x}{ucosA}=\dfrac{2musin\theta}{qE}\\\Rightarrow E=\dfrac{2mu^2sin\theta cos\theta}{xq}\\\Rightarrow E=\dfrac{2\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\times sin30\times cos30}{1.5\times 10^{-2}\times (-1.6\times 10^{-19})}\\\Rightarrow E=-1449.69404\ N/C$

The electric field is -1449.69404 N/C

8 0

3 years ago

Which of the following has a longer integration time (can collect light for a longer period

steposvetlana [31]

Answer:

d. all of the above (a through c)

Explanation:

Photographic film, Photographic plates and a charged couple device all have longer integration time than the human eye.

The eye optimizes the time interval during which light energy is accumulated until a picture is sent to the brain. This interval is called the "time of integration" and is typically 0.1 sec; for low-light levels, it increases to perhaps 0.2 sec.

8 0

3 years ago

A truck running at 115 km / h slows down at a rate of 4 m / s every 2 seconds. How far will it travel to a stop? choose the corr

Molodets [167]

Answer:

3) 255.04 m

Explanation:

Given:

v₀ = 115 km/h = 31.944 m/s

v = 0 m/s

a = -(4 m/s) / 2s = -2 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (31.944 m/s)² + 2 (-2 m/s²) Δx

Δx = 255.11 m

Closest answer is option 3.

4 0

3 years ago