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1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com

Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

$where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s$

$d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m$

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

$d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}$

$d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}$

$d' = 594\sqrt{1 - 0.81}$

$d' = 594 \times 0.4359$

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0

4 years ago

What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight ou

Lubov Fominskaja [6]

Answer:

The torque about his shoulder is 34.3Nm.

The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.

Explanation:

The solution to the problem can be found in the attachment below.

6 0

3 years ago

a man stands on weighing scale in a lift which carries him upwards with acceleration, what happen please tell

Lapatulllka [165]

He gets heavier due to the upward acceleration of the lift

5 0

4 years ago

What is the cell emf for the concentrations given? Express your answer using two significant figures.

alisha [4.7K]

Complete Question

A voltaic cell is constructed with two $Zn^{2+}$ $- Zn$ electrodes. The two cell compartment have $[Zn^{2+}] = 1.6 \ M$ and $[Zn^{2+}] = 2.00*10^{-2} \ M$ respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

The value is $E = 0.06 V$

Explanation:

Generally from the question we are told that

The concentration of $[Zn^{2+}]$ at the cathode is $[Zn^{2+}]_a = 1.6 \ M$

The concentration of $[Zn^{2+}]$ at the anode is $[Zn^{2+}]_c = 2.00*10^{-2} \ M$

Generally the the cell emf for the concentration is mathematically represented as

$E = E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}$

Generally the $E^o$ is the standard emf of a cell, the value is 0 V

So

$E = 0 - \frac{0.0591}{2} * log[\frac{ 2.00*10^{-2}}{1.6} ]$

=> $E = 0.06 V$

4 0

3 years ago

A bicycle has a momentum of 36 kg m/s and avelocity of 4 m/s. What is the mass of the bicycle?

miv72 [106K]

Momentum=36kgm/s
velocity=4m/s
mass=?
momentum=mass*velocity
mass=9kg

8 0

4 years ago

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