Can someone tell me the answer to this?

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th

Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0

3 years ago

if you dig a hole through the earth, from one hemisphere to the other, and drop a tennis ball into it, would it get stuck at the

tia_tia [17]

Answer:

Depending on which hemisphere it is, like western to eastern, It would most likely get stuck at the center. You would also have to put more things into thought like acceleration, velocity, and speed.

BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.

7 0

3 years ago

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I need someone smart!

Ivanshal [37]

Answer:

If we are looking for evidence of something that exists outside of our visible Universe and leaves no trace within it, it seems that the idea of a Multiverse is fundamentally untestable. But there are all sorts of things that we cannot observe that we know must be true. Decades before we directly detected gravitational waves, we knew that they must exist, because we observed their effects.

Explanation:

Maybe helps lol

5 0

3 years ago

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A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

$\mu =0.587$

So coefficient of friction will be 0.587

6 0

3 years ago

Technician A says the counter gear is actually several gears machined out of a single piece of steel. Technician B says the coun

dexar [7]

Answer:

Option c. (Both Technician A and B are correct)

Explanation:

A transmission system consists of 3 shafts. The input shaft, the counter shaft, and the main shaft. The clutch gear always rotates with input shaft and is a crucial element of the input shaft.

The counter shaft is actually several gears machined out of a single piece of steel. The counter shaft may also be called counter gear or cluster gear. It is a secondary shaft that runs parallel to the mainshaft in a gearbox and is used to provide powers to machine components such as the drive axle.

The main gears (also called the speed gears) on main shaft (also known as the output shaft) are used to transfer rotation from counter shaft to the output shaft.

Hence in the light of above description, both technician A and B are correct.

4 0

3 years ago

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