A car drives 20 miles north and then 3 miles south. What is the displacement

Which is true about carbon-12 compared with carbon-13

creativ13 [48]

E mass number of any given atom depends solely on the number of protons and neutrons in its nucleus. The mass number of any atom can be determined by adding the number of protons and neutrons. (Mathematically this is stated as Mass Number = Protons + Neutrons). For instance, a Carbon atom with 6 protons and 6 neutrons will have a mass number of 12AMU. However, a Carbon atom with 6 protons and 8 Neutrons will have a mass number of 14AMU. They are both Carbon atoms, however they each have a different mass number. Atoms of the same element that have different numbers of neutrons, and therefore, different mass numbers, are called isotopes. Isotopic symbols are used to indicate isotopes of the same elements. In the following isotopic symbols the lower number is the atomic number…it is the number of protons. The upper number is the mass number, it represents to sum of the protons and neutrons in the atoms nucleus.

7 0

2 years ago

A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini

finlep [7]

<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁, object 2 be u₂

Final velocity object 1 be v₁, object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

Initial momentum = Final momentum

24 = 3v₁ + 6M

v₁ + 2M = 8

v₁ = 8 - 2M

Initial kinetic energy = Final kinetic energy

96 = 1.5 v₁² + 18 M

v₁² + 12 M = 64

Substituting v₁ = 8 - 2M

(8 - 2M)² + 12 M = 64

64 - 32M + 4M² + 12 M = 64

4M² = 20 M

M = 5 kg

Option 2 is the correct answer.

6 0

2 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

Generally the length of metre is equal all over the world why?

erastova [34]

Answer:

It is an SI unit

Explanation:

The metre is defined as the length of the path travelled by light in a vacuum in 1299 792 458 of a second. The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole

8 0

3 years ago

The scientist who suggested that energy can be created under certain conditions was

tester [92]

Einstein hope this helped

7 0

3 years ago

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