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Yakvenalex [24]

3 years ago

10

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

answer in seconds, minutes, and hours

1 answer:

sp2606 [1]3 years ago

4 0

We have that the time in seconds, minutes, and hours is

$t=1.083*10^{-4}s$

$T_{min}=1.805*10^{-6}min$

$T_{hours}=3.0083*10^{-8}hours$

From the Question we are told that

Velocity $v=13.0m/s$

Distance $d=120 km$

Generally the equation for the Time is mathematically given as

$t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s$

Therefore

$T_{min}=1.083*10^{-4}s/60$

$T_{min}=1.805*10^{-6}min$

And

$T_{hours}=T_{min}/60$

$T_{hours}=3.0083*10^{-8}hours$

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when two resistors are wired in series with a 12 v battery, the current through the battery is 0.31 a. when they are wired in pa

liberstina [14]

Let R₁ and R₂ be the two Resistance

R₁ - Resistance of 1st resistor

R₂ - Resistance of 2nd resistor

V = Voltage = 12V

I = O.31A ( in series)

I = 1.6 A ( in parallel)

when two resistance connected in series

Rs = R₁ + R₂

V = IRs = 12 /0.31 V

R₁+R₂ = 38.700Ω (equation1.)

When two resistance connected in parallel

$R_{p} =\frac { R_{1} R_{2}}{R_{1} + R_{2} }$

V = I Rp

V= I (R₂ R₁ /R₁ +R₂)

R₁ R₂ = 290.25 Ω

As we know

(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2

Using above values we get,

(R₁ - R₂ ) ² = (38.70) ² - 4x 290.25

(R₁ - R₂ ) ² = 336.69 Ω

R₁ - R₂ = 18.34 (equation 2 )

Using equation 1 and 2 we get

R₁+ R₂ = 38.70 52

R₁ - R₂ = 18.34

R₁ = 28.535 Ω

Using the value of R₁ in equation 1 we get

R₂ = 38·700 - 28.535

RR₂ = 10.125 Ω

R₁ = 28. 535 v and R₂ 10.125 Ω

To know more about resistance in series and parallel:

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7 0

1 year ago

Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro

elena-14-01-66 [18.8K]

Solution :

Energy of photon, E = 6.7 eV

E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

$=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

= 0.00000000933

Relative mass, $m_{rel}=r.m$

$=7131 \times 1.6728 \times 10^{-27}$

$=1.1927 \times 10^{-23}$ kg

6 0

3 years ago

9. The acceleration due to gravity on the

laiz [17]

W=mg
W=75(1.6)
W=120N
a. 120N

6 0

3 years ago

A VW beetle goes from zero to 27m/s in 7.6 seconds. What is the acceleration?

Firlakuza [10]

Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2 (3 s.f)

7 0

3 years ago

If the resistivity of copper is less than that of gold at room temperature, which of the following statements must be true? Gold

KiRa [710]

Answer:

Gold Has A Higher Resistance Than Copper. The Sample Of Gold Is Thinner Than The Sample Of Copper. Electrons In Gold Are More Likely To Be Scattered Than Electrons In Copper At Room Temperature When they are exelerated by the same electric field.

Explanation:

5 0

3 years ago