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tester [92]
3 years ago
15

Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of v=0.8c

and determines its lifetime to be τA=129 ns. Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory?
Physics
1 answer:
Studentka2010 [4]3 years ago
3 0

Answer:

30.96 m

Explanation:

If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:

d = v * Δt

v = 0.8*c = 0.8 * 3e8 = 2.4e8

Δt = ζ = 129 ns = 1.29e-7 s

d = 2.4e8 * 1.29e-7 = 30.96 m

This is the distance as measured by observer A.

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NNADVOKAT [17]

Answer:

letter C. velocity hope this helps

7 0
3 years ago
A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

4 0
2 years ago
It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd tim
GaryK [48]

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

6 0
3 years ago
Write the device modeling equation for the first and second law of thermodynamics
anyanavicka [17]
1st <span>the total </span>energy<span> of an </span>isolated system<span> is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. ▲U=Q-W
</span><span> 
2nd the total </span>entropy<span> can never decrease over time for an </span>isolated system, that is, a system in which neither energy nor matter can enter nor leave.
DS (Greater than or equal to) 0
8 0
3 years ago
An object is pulled northward with a force of 10 n and southward with a force of 15 n. the magnitude of the net force on the obj
Anastaziya [24]

The correct option is (b) 5n

As a result, there is a net downward force of 5N operating on the object.

The resultant force is the force that results from adding the vector sums of all the forces operating on an item. The combined action of all the acting forces on the object produces the same effect as the resulting force. When determining the resulting force, the direction of the forces must be taken into account.

Given;

The northward force is Fn = 10N

The southward force is Fs = 15N

Required;

The net force on the mobile phone is Fnet = ?N

The object's weight exerts downward pressure, and upward resistance exerts upward pressure. The vector sum of these two forces will be the net force.

Fnet = Fs - Fn (Considering the direction downward as positive)

Fnet= 15N - 10N

Fnet = 5N

As a result, there is a net downward force of 5 N operating on the object.

Learn more about the Force with the help of the given link:

brainly.com/question/7362815

#SPJ4

6 0
1 year ago
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