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3 years ago

Design a roller coaster.

Physics

1 answer:

Zarrin [17]3 years ago

8 0

1. Because the cart on the coaster must get all of its power from this first hill.
2. It must be smaller because if it was bigger the cart would roll backwards without enough power.
3. an upward climb that is too steep.

I answered the first 3 but try to do the last two by yourself. They are very easy once you figure it out. Try searching the definition of the main points. :)

You might be interested in

An antibaryon composed of two antiup quarks

Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have $+\frac{2}{3} e$ charge where as down, strange and bottom quarks have $-\frac{1}{3}e$ charge.

The antiparticle of up quark is antiup quark and has charge $-\frac{2}{3}e$ charge.

The antiparticle of down quark is antidown quark and has charge $+\frac{1}{3}e$ charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

$2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e$

Thus, antibaryon has -1e charge.

5 0

4 years ago

A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?

katrin [286]

Search Results<span>By simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

Hope This Helps!</span>

7 0

3 years ago

This engineer invented the iron wire to replace hemp rope.

IceJOKER [234]

The answer is A John Roebling.

7 0

4 years ago

Read 2 more answers

Ivahew [28]

Explanation:

Speed = distance / time

v = 560 m / 25 s

v = 22.4 m/s

8 0

4 years ago

Star a has an absolute magnitude of 8. what is its apparent magnitude at a distance of 100 pc

tresset_1 [31]

From the definition of apparent magnitude, we know that:
$m_{1} - m_{2} = -2.5Log( \frac{F_{1}}{F_{2}} )$
where:
m = apparent magnitude
F = corresponding flux

We also know that the flux is given by the formula:
$F = \frac{L}{4 \pi d^{2} }$
where:
L = luminosity
d = distance

Therefore:
$\frac{F_{1} }{F_{2}} = (\frac{L_{1} }{4 \pi d_{1}^{2} })(\frac{4 \pi d_{2}^{2} }{L_{2} }) \\ = \frac{L_{1}d_{2}^{2}}{L_{2}d_{1}^{2}}$

Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
$m - M = 2.5 Log ( \frac{d}{10pc})^{2}$

Now, let's solve for m:
$m = M + 2.5 Log ( \frac{d}{10})^{2}$
= <span> $8 + 2.5 Log ( \frac{100}{10})^{2}$ </span>
= 13

Hence, the apparent magnitude of the star would be m = +13

3 0

4 years ago