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3 years ago

Which statement accurately describes CI2?

2 answers:

8 0

1. 100,000% A. I simply googled C12 and it's the most stable Isotope of Carbon...

2. A as well.. 99% sure.. and so far my 99% sure has been 100% correct so ... :p

vlada-n [284]3 years ago

3 0

Uhhhh Vote Hillary For Lethal Injection...

Bush did 9/11

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The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a

diamong [38]

Answer: $\theta =9.96^{\circ}$ North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus $195sin\theta =33.9$

$\theta =9.96^{\circ}$

so Plane should head towards $9.96^{\circ}$ North of west in order to fly in west.

So plane

actual velocity is

$v=-195cos9.96\hat{i}+195sin9.96\hat{j}$

5 0

3 years ago

Which of the following is a correct definition of radiant energy?

ValentinkaMS [17]

Radiant energy is the energy of electromagnetic and gravitational radiation

6 0

2 years ago

Read 2 more answers

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

3 years ago

A light-year is _______________

kozerog [31]

The correct answer would be A "<span>A light-year is the distance light travels in a year.

This is considered a unit of distance connected to the distance that light can travel in one year. It is proved that light travels at 300,000 km per second so, in 1 year, it might travel 10 trillion km.

</span>

3 0

3 years ago

Read 2 more answers

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$