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attashe74 [19]
3 years ago
11

John rides his bike with a constant speed of 12 miles per hour. How far can he travel in 2 hours?

Physics
1 answer:
Gwar [14]3 years ago
5 0

Answer:

24 miles

Explanation:

Make a proportion

He is traveling at a constant rate of 12 miles/hour, so

he can go 12 miles in 1 hour, and x miles in 2 hours

12/1=x/2

Multiply both sides by 2

2*12/1=x/2 *2

24=x

So, in 2 hours he can travel 24 miles

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g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell
olga55 [171]

Answer:

The value is V  =  2 V

Explanation:

From the question we are told that

The length of the wire is l = 10 \ m

The current density is J =  4*10^6 \  A/m^2

The conductivity is \sigma  =  2*10^{7} \  S/m

Generally conductivity is mathematically represented as

\sigma  =  \frac{l}{RA}

Here R is the resistance which is mathematically represented as

R =  \frac{V}{I}

Here I is the current which is mathematically represented as

I  =  J * A

So

R =  \frac{V}{  J * A}

And

\sigma  =  \frac{l}{\frac{V}{  J * A} * A}

=> \sigma  =  \frac{l}{\frac{V}{J}}

=> V = \frac{l * J}{\sigma }

=> V = \frac{10  * 4*10^6}{2*10^{7}  }

=> V  =  2 V

5 0
3 years ago
The blue mass is currently 4 meters away from the red mass.
umka21 [38]

Answer:

its a!

Explanation:

•✧•

6 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
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