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Akimi4 [234]
4 years ago
11

Which of the following could be used to sterilize plastic petri plates in a plastic wrapper? A) ultraviolet radiation B) microwa

ves C) autoclave D) sunlight E) gamma radiation
Physics
1 answer:
Ivahew [28]4 years ago
8 0

Gamma radiation could be used to sterilize plastic petri plates in a plastic wrapper.

Answer: Option E

<u>Explanation: </u>

It is known that radioactive radiations have greater ionizing powers compared to other electromagnetic radiations. Among the radioactive radiations, gamma rays are the most ionizing radiation.

So they are widely used in sterilizing process. As the penetration power of gamma is more compared to alpha and beta, in order to kill microorganisms gamma radiations are used.

When plastic petri plates wrapped with plastic are bombarded or sterilizing with gamma rays, the rays will ionise the micro-organism. Thus mutilating their DNA and preventing their reproduction.

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I'm positive it's the last one

Explanation:

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The________ channels sound to the eardrum.
inn [45]
The external auditory meatus, or ear canal, funnels sound to the eardrum. It is part of the outer ear.
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(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational
Oksanka [162]

Answer:

Explanation:

a) using the energy conservation  equation

mgh = 0.5mv^2 + 0.5Iω^2

I(moment of inertia) (basket ball) = (2/3)mr^2

mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)

gh = 1/2v^2 + 1/3v^2

gh = v^2( 5/6)

v =  \sqrt{\frac{6gh}{5} }

putting the values we get

6.6 ^{2} = \frac{6\times9.8h}{5}

solving for h( height)

h = 3.704 m apprx

b) velocity of solid cylinder

mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)

g*h = 1/2v^2 + 1/4v^2

g*h = 3/4v^2

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Irina18 [472]
The mass of an object divided by its volume
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A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above
babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

h(t) = h_0 + ut - \frac{1}{2}gt^2 (1)

where

h_0 is the height at time t = 0

u is the initial vertical velocity

g=32 ft/s^2 is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

f(t)=-16t^2+44t+88 (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

h_0 = 88 ft

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

f(t)=0

which means when

0=-16t^2+44t+88

First of all, we can simplify the equation by dividing each term by 4:

0=-4t^2+11t+22

This is a second-order equation, so we solve it using the usual formula and we find:

t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

v=u-gt

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

v=0

which means

0=u-gt

Here we have

u=+44 ft/s (initial velocity)

g=32 ft/s^2 (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

t=\frac{u}{g}=\frac{44}{32}=1.38 s

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

f(t)=-16t^2+44t+88

Substituting t = 1.38 s, we find:

f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m

So, the maximum height reached by the rock during its motion is

h_{max}=118.2 m

Which means 118.2 m above the water.

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4 years ago
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