1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
hoa [83]
2 years ago
6

A beam of unstable K mesons, traveling at a speed of 32c, passes through two counters 9.00 m apart. The particles suffer a negli

gible loss of speed and energy in passing through the counters but give electrical pulses that can be counted. It is observed that 1000 counts are recorded in the first counter and 250 in the second. Assuming that this whole decrease is due to the decay of the particles in flight, what is their half-life as measured in their own rest system?
Physics
1 answer:
Anna71 [15]2 years ago
5 0

Answer:

8.66 nano seconds

Explanation:

speed of k mesons ( v ) = c \sqrt{3} /2  m/s

Distance between counters ( d ) = 9.00 m

number of countable electrical pulses = 1000 counts in first counter and 250 in second counter

time of travel = d / v  = 18 / \sqrt{3} C  secs

Next write the decay of particles in lab frame

finally calculate the half life of Meson in its own frame

( t 1/2) of meson in its own frame = 8.66 n-secs

attached below is a detailed solution

You might be interested in
Which description explains the flow of heat?
Lemur [1.5K]

Answer:

the answer is b

Explanation:

7 0
3 years ago
Read 2 more answers
The speed of a projectile when it reaches its maximum height is 0.58 times its speed when it is at half its maximum height. What
Ne4ueva [31]

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

v_1 = v_x

now when he is at half of the maximum height the speed will be in x and y direction both

v_2 = \sqrt{v_y^2 + v_x^2}

here it is given that

v_1 = 0.58 v_2

v_x = 0.58\sqrt{v_x^2 + v_y^2}

2.97 v_x^2 = v_x^2 + v_y^2

1.97 v_x^2 = v_y^2

also we know that

v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}

here we know that maximum height is given as

H = \frac{v_{iy}^2}{2g}

v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}

v_y^2 = \frac{v_{iy}^2}{2}

now from above

1.97 v_x^2 = \frac{v_{iy}^2}{2}

1.98 v_x = v_{iy}

also we know that angle of projection is

tan\theta = \frac{v_{iy}}{v_x}

tan\theta = \frac{1.98v_x}{v_x}

so angle is

\theta = tan^{-1} 1.98

\theta = 63.3 degree

6 0
3 years ago
A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj
LenaWriter [7]

Answer:

485520 m

Explanation:

v_{o} = initial velocity of the projectile = 1360 m/s

v_{f} = final velocity of the projectile = \left ( \frac{2}{5} \right )v_{_{o}} = \left ( \frac{2}{5} \right )(1360) = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a h

Inserting the values

544^{2} = 1360^{2} + 2 (-1.6) h

h = 485520 m

3 0
3 years ago
In a skating stunt known as "crack-the-whip," a number of skaters hold hands and form a straight line. They try to skate so that
4vir4ik [10]

Answer:

The magnitude of the centripetal force that acts on him

Explanation:

Given that,

Mass = 80.0 kg

Distance = 6.10 m

Speed = 6.80 m/s

We need to calculate the magnitude of the centripetal force that acts on him

Using formula of the centripetal force

F_{c}=\dfrac{mv^2}{r}

Where, F = force

m = mass

v = speed

r = distance

Put the value into the formula

F_{c}=\dfrac{80.0\times(6.80)^2}{6.10}

F_{c}=606.4\ N

Hence, The magnitude of the centripetal force that acts on him

7 0
3 years ago
The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th
Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec,  B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>

at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 =  180 rad/sec

<u>b) Determine the force exerted on B2 at P</u>

T2 = mB1g + T1  -------- ( 1 )

where ; T1 = mB2g  ( at point p )

                 = 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

<u />

3 0
3 years ago
Other questions:
  • Why did scientists using classical, Newtonian physics have difficulty explaining the photoelectric effect?
    9·2 answers
  • April swims 50m north and then turns and swims back to the south. What is her velocity if it takes 60 seconds
    5·1 answer
  • What is the final velocity of a 10 kg object that drops from rest and falls for 30 seconds?
    13·1 answer
  • A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its o
    8·1 answer
  • "Nuclear stability is based on (choose
    5·1 answer
  • Elements are composed of atoms of at least two different types.<br> True<br> False
    8·1 answer
  • When does the edge of the water advance farther inland- when the ocean's floor is a gradual slope or steep slope?
    9·1 answer
  • Which of the following is an example of a healthy behavior?
    10·1 answer
  • What is third law of motion​
    15·2 answers
  • A diving board of length 3.00 mm is supported at a point 1.00 mm from the end, and a diver weighing 550 NN stands at the free en
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!