Which of the following is not a compound A. Gold B. Water C. Salt D. Sugar

Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of

kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0

3 years ago

Convertir 1200 ms a cs Convertir 0,3 mm a cm.

Vaselesa [24]

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

3 0

3 years ago

Question 1(Multiple Choice Worth 4 points) The star named Canopus has a declination of approximately –52°. Which of these statem

Y_Kistochka [10]

Answer:

It is 52° below the celestial equator.

Explanation:

The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.

The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary

The star named Canopus has a declination of approximately –52°.

Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.

Thus, the star named Caponus is 52° below the celestial equator.

8 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5t, collide with molecules of t

andreev551 [17]

The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

4 0

3 years ago

A space traveller leaves Earth for 10 years at .85c. According to an observer on Earth, how much time has passed?

eduard

First of all, you didn't tell us WHO measured the "10 years".

If it was the people on Earth, then 10 years passed according to them.

If it was 10 years on the space traveler's clock, then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.

If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by

√(1 - v²/c²) .

You said that v/c = 0.85 .

v²/c² = (0.85)² = 0.7225

1 - v²/c² = 1 - 0.7225 = 0.2775

√(1 - v²/c²) = √0.2775 = 0.5268

If one clock counts up 10 years, then the other one counts up

(10years) / 0.5268 = 18.983 years 

I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.

8 0

3 years ago