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Natalija [7]
3 years ago
7

Two charged particles move in the same direction with respect to the same magnetic field. particle 1 travels two times faster th

an particle 2. however, each particle experiences a magnetic force of the same magnitude. find the ratio q1/q2 of the magnitudes of the charges.
Physics
1 answer:
MaRussiya [10]3 years ago
6 0
<span>0.166666666667

Hope this helped!
STSN</span>
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Create a list of at least five different shapes of signs. For each of the five items on your list, identify two specific signs.
Nitella [24]
Hey bud, Good to see you again!

Okay so lets get to work huh? :D

Circle - Railroad advance warnings, and some bus signs.
Octagon - Stop signs, and slow signs (for flooding on roads).
Triangle - People crossing, People working.
Square - parking, speed limit.
Diamond - Bicyclist, animal crossing.

Hope this helped. Have a great day!


5 0
4 years ago
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Can someone summarize the multiverse theory? I need it for a quick homework assignment :]
Whitepunk [10]

Answer:

The multiverse is a hypothetical group of multiple universes. Together, these universes comprise everything that exists: the entirety of space, time, matter, energy, information, and the physical laws and constants that describe them.

Hope it helped you :D

5 0
3 years ago
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Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

4 0
3 years ago
If you took a 20 g rock from the Earth to the moon, which would change?
Fudgin [204]

Answer:

weight of tock would change

6 0
4 years ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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