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sertanlavr [38]
3 years ago
12

Which of the following is an example of pure research? Creating new elements to study their properties Producing plastics that a

re heat-resistant to use in ovensDetermining how to reduce rusting on metal ships and cargo carriersFinding alternatives to gasoline to fuel transportation vehicles
Physics
1 answer:
koban [17]3 years ago
6 0
Creating new elements in order to study their properties is definitely pure research.
It has no earthly [known] application, since the elements don't even exist, whereas
for each of the other choices, you've stated the desired application as the reason
for the search.
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Which of the following is NOT a characteristic of noble gases?
Rudik [331]
Solid at room temperature
8 0
2 years ago
Read 2 more answers
A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The
IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0
3 years ago
Kon'nichiwa~<br>please help me with this question!!​
andrezito [222]

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

4 0
2 years ago
What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?
Vlada [557]

Answer:

Ep=-2.3*10^{-19}J

Explanation:

The change in potential energy can be expressed as:

Ep=K.\frac{q1.q2}{r}

where K is a constant with a value of 9*10^{9}\frac{N.m^{2}}{C^{2}}, q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is +1.6*10^{-19}C and the charge for the electron is -1.6*10^{-19}C.

Converting r=1.0nm to m:

1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m

Replacing values:

Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}

Ep=-2.3*10^{-19}J

5 0
2 years ago
An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the
Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image​. Let v be the image distance. Using lens formula to find it :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put all the values,

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm

So, the image distance from the lens is 24 cm.

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2

The negative sign of magnification shows that the formed image is inverted.

7 0
2 years ago
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