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3 years ago

What is one positive outcome of the recentevents?

Physics

2 answers:

inna [77]3 years ago

6 0

The nhl because you have no other

fenix001 [56]3 years ago

5 0

Answer:

Analyze the positive and negative consequences of catastrophic events of the last 40-50 years and the individuals that had an impact on these events. Meaning.

Explanation:

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Partial. a penumbral lunar eclipse is when the moon passes through the earths penumbra

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3 years ago

What simple machines combine to make a shovel which is a compound

IRINA_888 [86]

B) lever and pulley is the answer

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2 years ago

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$

Therefore, the particle's charge-to-mass ratio is $958.1\times10^5C/kg$

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

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3 years ago

Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe

solniwko [45]

"v0" means that there are no friction forces at that speed
mgsinΘ = (mv0²/r)cosΘ → the variable m cancels 
sinΘ/cosΘ = tanΘ = v0² / gr
Θ = arctan(v0² / gr) 

When v > v0, friction points downslope: 
mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: 
gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ 
µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) 
where Θ is defined above. 

When v > v0, friction points upslope: 
mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: 
gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ 
µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) 
where Θ is defined above.

4 0

3 years ago

What are the 3 formulas you can use for vertical motion for a projectile?

klio [65]

I hope this can help you ask me if you need help again

4 0

3 years ago

What is one positive outcome of the recentevents?​

What is one positive outcome of the recentevents?