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Sav [38]
3 years ago
5

which type of lightening device would you use for each of the following needs: an economical light source in a manufacturing pla

nt, an eye-catching sign that will be visible at night, and a basketball stadium? Explain - PHYSICAL SCIENCE
Physics
1 answer:
Yakvenalex [24]3 years ago
7 0
Btw only someone who is nice will answer tour question. You can't expect for explanition when the question is only worth 5 points. Not trying to be mean sorry if i am being mean
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Power = (240 volts) x (4 Amp)

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Can You Please Give Me Some Examples of Density-Independent and Density-Dependent Limiting Factors?
ivolga24 [154]
Density-Dependent: 
1<span><span><span><span>. </span>competition.</span><span>   
<span>2. </span>overcrowding.</span><span>   
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3 years ago
If the distance between two asteroids is doubled, the gravitational force they exert on each other will.
True [87]
The force will be 4 times smaller.
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3 years ago
How does inertia affect someone who isn't wearing a seatbelt
Paul [167]
Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
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7 0
3 years ago
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.
Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

d=30,000 pc =9.26\cdot 10^{20} m

The speed of the cosmic ray is

v=0.98 c

where

c=3.0 \cdot 10^8 m/s is the speed of light. Substituting,

v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s

Converting into years,

T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}

And substituting v = 0.98c, we find:

T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years

3 0
3 years ago
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