Which group of elements loses 1 electron easily ?

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

4 years ago

A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o

Rashid [163]

a) 1.57 m/s

The sock spins once every 2.0 seconds, so its period is

T = 2.0 s

Therefore, the angular velocity of the sock is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s$

The linear speed of the sock is given by

$v=\omega r$

where

$\omega$ is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

$v=(3.14)(0.50)=1.57 m/s$

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

$r' = 2r = 1.00 m$

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

$\omega' = \omega = 3.14 rad/s$

Therefore, the new linear speed would be:

$v'=\omega' r' = \omega (2r)$

And substituting,

$v'=(3.14)(1.00)=3.14 rad/s = 2v$

So, we see that the linear speed has doubled.

8 0

3 years ago

You are walking from your math class to your science class. You are carrying books

kolezko [41]

Answer:

1800J

Explanation:

Given parameters:

Weight of the book = 20N

Total distance covered = 45m + 15m + 30m = 90m

Unknown:

Total work performed on the books = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

Work done = Force x distance

Work done = 20 x 90 = 1800J

8 0

4 years ago

If an athlete expends 3480. kJ/h, how long does she have to play to work off 1.00 lb of body fat? Note that the nutritional calo

Sunny_sXe [5.5K]

Answer:

The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

Explanation:

1 lb of body fat = 4.1 k cal

1 k cal = 4.184 Kilo joule

1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule

Athlete expends 3480 Kilo joule in one hour

⇒ Time required to expand 3480 Kilo joule = 60 minute

⇒ Time required to expand 1 Kilo joule = $\frac{60}{3480}$ $\frac{min}{KJ}$

⇒ Time required to expand 17.1544 Kilo joule = $\frac{60}{3480}$ × 17.1544 = 0.296 min

Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

6 0

3 years ago

What is the middle layer of the sun atmosphere

lana [24]

The chromosphere is the middle layer

7 0

4 years ago

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