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iragen [17]
1 year ago
11

An astronaut travels to a star system 3.9 lyly away at a speed of 0.90 cc . Assume that the time needed to accelerate and decele

rate is negligible.
Physics
1 answer:
kakasveta [241]1 year ago
4 0

Total time elapsed is =8.2y

The starting event is the astronaut leaving Earth. The finishing event is the astronaut arriving at the star system. The time between these events on Earth is:

Δt=3.9ly/0.9c

Δt=4.3y

For the astronaut, two events occur at the same position and can be measured with just one clock. Hence,

Δτ

=  \sqrt{1 -  \frac{v {}^{2} }{c {}^{2} }  }  \times Δt

Δτ

=  \sqrt{1 - ( \frac{0.9c}{c} ) {}^{2} }  \times (4.3y)

Δτ=1.8ly

The total elapsed time is:

T elapsed=Δt+3.9

T elapsed=4.3+3.9

T elapsed=8.2y

learn more about time from here: brainly.com/question/28208983

#SPJ4

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Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho
Jlenok [28]

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

3 0
3 years ago
Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
3 years ago
A cylindrical tungsten filament 14.0 cmcm long with a diameter of 1.00 mmmm is to be used in a machine for which the temperature
Advocard [28]

Answer:

Resistivity ρ=1.12 x 10^-4 Ωm

Explanation:

ρ= RA/l, where R is resistance, A is cross sectional area and l is length

A=πr^2

Note Current is given R is proportion to temperature and inversely proportional to Current R=(20+273)/14*10^-2 =2000Ω

⇒ρ=R*πr^2/l all length in metre.

8 0
3 years ago
4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
11111nata11111 [884]

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

3 0
3 years ago
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