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2 years ago

How long will it take you to pass a truck at 60 mph with oncoming traffic?

Physics

2 answers:

DedPeter [7]2 years ago

6 0

<span>This is impossible to calculate without knowing the speed of each vehicle. </span>

kobusy [5.1K]2 years ago

5 0

Answer:

4 seconds - Not practical

Explanation:

Length of the truck = 50'

Initial distance behind the truck = 30'

Finish Pass = 50' ahead of truck ,

Pass at = 60mph -- about 3.375 seconds.

- 70mph your closing speed is 130mph.

If you were less than a 1/4 mile away when you tried the pass you will be dead.

That would be a quick pass. You will probably want a mile beyond the oncoming traffic.

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Why do adults make bigger splashes when they jump into swimming pools than small children?

Semenov [28]

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

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2 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

2 years ago

What do we call magnets that are only magnetized when they are within an existing magnetic field?

Zigmanuir [339]

Answer: I think the answer is D

Explanation: N/A

3 0

2 years ago

Please help me find the equations guys

Alexeev081 [22]

The line at the bottom of the picture ... probably the first line on a list of choices .. is the correct equation.

4 0

2 years ago

What is the magnitude of the electric force on an electron in a uniform electric field of strength 2050 n/c that points due east

erastova [34]

Given E=2050 N/C

charge of an electron = 1.6×10^-19 C

USING E = F/Q , electric field = force ÷ charge

2050 = F ÷ (1.6×10^-19)....

make F subject of formula and get the answer

3 0

3 years ago