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3 years ago

Use the collision theory to explain how increasing the temperature of a reaction will affect the rate of the reaction.

1 answer:

7 0

Increasing the temperature causes the particles in the reaction to become kinetically excited, hitting one another in increasing frequency. Increased collision among means faster rate or reaction.

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A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

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3 years ago

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1. The hockey puck is given an initial speed of 20.0m/s on a frozen pond. The

andreyandreev [35.5K]

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3 years ago

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An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

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3 years ago

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The peak luminosity of a white dwarf supernova is around 1010 Lsun, and it remains brighter than 108 Lsun for about 150 days. In

Airida [17]

Answer:

Explanation: find the attached solution below

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3 years ago