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GaryK [48]
3 years ago
13

A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt

er it passes through. One such particle, initially traveling at 2.40 ✕ 106 m/s in a straight line, decreases in speed to 1.56 ✕ 106 m/s over a distance of 1.22 km.
(a) What is the magnitude of the force experienced by the muon?
(b) How does this force compare to the weight of the muon?
|F|/Fg =______
Physics
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

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as you go above earths surface, the acceleration due to gravity will decrease. find the height, i meters, above the earths surfa
aleksklad [387]

Answer:

1.23×10⁸ m

Explanation:

Acceleration due to gravity is:

a = GM / r²

where G is the universal gravitational constant,

M is the mass of the planet,

and r is the distance from the center of the planet to the object.

When the object is on the surface of the Earth, a = g and r = R.

g = GM / R²

When the object is at height i above the surface, a = 1/410 g and r = i + R.

1/410 g = GM / (i + R)²

Divide the first equation by the second:

g / (1/410 g) = (GM / R²) / (GM / (i + R)²)

410 = (i + R)² / R²

410 R² = (i + R)²

410 R² = i² + 2iR + R²

0 = i² + 2iR − 409R²

Solve with quadratic formula:

i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)

i = [ -2R ± √(1640R²) ] / 2

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i = -R ± R√410

i = (-1 ± √410) R

Since i > 0:

i = (-1 + √410) R

R = 6.37×10⁶ m:

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8 0
3 years ago
Car A is traveling west at 40 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two road
Gwar [14]

Explanation:

It is given that,

    \frac{dx}{dt} = -40 mi/h,     \frac{dx}{dt} = -40 mi/h

The negative sign indicates that x and y are decreasing.

We have to find \frac{dz}{dt}. Equation for the given variables according to the Pythagoras theorem is as follows.

              z^{2} = x^{2} + y^{2}

Now, we will differentiate each side w.r.t 't' as follows.

        2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}

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So, when x = 4 mi, and y = 3 mi then z = 5 mi.

As,       \frac{dz}{dt} = \frac{1}{z}(x\frac{dx}{dt} + y\frac{dy}{dt})

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