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3 years ago

How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

1 answer:

5 0

Well, I may not know the exact answer, but I can confirm because of mass. The more mass an object has, the less control you have on it, so it can't be stopped easily. Let's take a small rock and a boulder for example. When a small rock rolls on the ground, it can easily be stopped by your hand or foot. But when you have to deal with an enormous boulder, you can't easily stop it with one just person. You'll need more people. So the answer is because the smaller vehicle has a lesser mass than the larger vehicle. (if you are looking for the answer, just look at the end.

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If vector A=2i+3j-k and vector B=4i+6j-2k. The angle between them will be: a) 0°b)45°c)90°d)60°

melomori [17]

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3 years ago

Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy

Dominik [7]

a) 0.261 T

b) this field strength is obtainable with today's technology

Explanation:

The force experienced by a charged particle moving perpendicular to a magnetic field is given by

$F=qvB$

where

q is the charge

v is the velocity

B is the strength of the field

This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the circle

In this problem, we have:

$q=1.6\cdot 10^{-19}C$ (magnitude of the charge of antiprotons)

$v=5.00 \cdot 10^7 m/s$ (velocity)

$m=1.67\cdot 10^{-27}kg$ (mass of antiprotons)

r = 2.00 m (radius)

Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:

$B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T$

The strength of the magnetic field calculated in part A) is

$B=0.261 T$

This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about

$B_{earth}=5\cdot 10^{-5} T$

However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).

Therefore, we can say that this field strength is obtainable with today's technology.

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3 years ago