If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered


Answer 1


109.5 million years


The question asked us to find the time.

Remember that

Rate of velocity = distance / time, and this,

time taken = distance/rate

Due to the confusing nature of the units, we would have to be converting them to a more uniform one.

1 km is equal to 9.461*10^12 km/light-year, that's if we try to convert km to light year.

Since the speed is in km, the distance has to be in km also, and therefore, we convert ly to km:

4.5 light-years = 9.461*10^12 km/light-year) = 42.57*10^13 km

We that this value as our distance, in km.


Time = distance/speed

Time = 45.57*10^13 km / 490 km/hr = 9.3*10^11 hr

Now the next step is to convert hours to years, using the conversion factor 8766 hr/yr.

time (in years) = 9.6*10^11 hr / 8766 hr/yr) = 10.95*10^7 years

the final step is to divide the time in years by 10^6 years/million years, which gives the final answer as the trip takes 109.5 million years.

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A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing? b. is the electric field in the direction of the current or opposite to the current?

A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.



It is given that,

Mass of the tennis ball, m_1=0.06\ kg

Initial speed of tennis ball, u_1=5.28\ m/s

Mass of ball, m_2=0.08\ kg

Initial speed of ball, u_2=3\ m/s

In case of elastic collision, the momentum remains conserved. The momentum equation is given by :


v_1\ and\ v_2 are final speed of tennis ball and the ball respectively.

0.06* 5.28+0.08* 3=0.06v_1+0.08v_2


We know that the coefficient of restitution is equal to 1. It is given by :




On solving equation (1) and (2) to find the values of velocities after collision.

v_1=5.28\ m/s

v_2=3\ m/s

So, the speed of both balls are 5.28 m/s and 3 m/s respectively. Hence, this is the required solution.

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)





We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)


Lol I would help you but I have no clue

1. A bicyclist starts at 2.5 m/s and accelerates along a straight path to a speed of 12.5 m/s ina time of 4.5 seconds. What is the bicyclist's acceleration to the nearest tenth of a m/s??








The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel wire with this length and a cross-sectional area of 3.10 ✕ 10^−6 m^2 is laid on the bridge deck with its ends attached to the towers of the bridge, on a summer day when the temperature of the wire is 43.0°C. When winter arrives, the towers stay the same distance apart and the bridge deck keeps the same shape as its expansion joints open. When the temperature drops to −10.0°C, what is the tension in the wire? Take Young's modulus for steel to be 20.0 ✕ 10^10 N/m^2. (Assume the coefficient of thermal expansion of steel is 11 ✕ 10−6 (°C)−1.)



361.46 N


\alpha = Coefficient of thermal expansion = 11* 10^(-6)\ /^(\circ)C

Y = Young's modulus for steel = 20* 10^(10)\ Pa

A = Area = 3.1* 10^(-6)\ m^2

L_0 = Original length = 1.28 km

\Delta T = Change in temperature = 45-(-10)

Length contraction is given by

\Delta L=\alpha L_0\Delta T


\Delta L=(L_0T)/(YA)

\alpha L_0\Delta T=(L_0T)/(YA)\n\Rightarrow T=\alpha \Delta TYA\n\Rightarrow T=11* 10^(-6)* (43-(-10))* 20* 10^(10) * 3.1* 10^(-6)\n\Rightarrow T=361.46\ N

The tension in the wire is 361.46 N

For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?


Answer: ∆p2 = 2* ∆p1


Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L


∆p2/L2 = ∆p1/L1

Since L2 = 2 * L1

∆p2/2*L1 = ∆p1/L1

Eliminating L1 we have,

∆p2/2 = ∆p1

Multiplying both sides by 2

∆p2 = 2 * ∆p1