does the sabre-tooth currirulum still exist at present?give examples of your evidence​

Answer:

Eliza’s suggestion is the most promising

Explanation:

Newton’s law of universal gravitation is [tex]F = G\frac{m_1m_2}{r^2}[/tex] where G is a constant and both masses are constant in this experiment as well. So it only depends on r, the distance between the center of mass of both objects.

Bringing it high in the sky, as Flo suggests, is definitely not a good idea because it only increases the distance from the center of the Earth mass.

A mountain contains significant mass, so the center of the Earth mass is somewhat shifted to regions with the largest mountains. However, standing on top of a mountain, as Lorenzo suggests, doesn’t help since the shift of the center of mass, if any, is far smaller than the height of the mountain.

Standing near sea level, as Eliza suggests, is a good way to minimize the distance to the center of the Earth mass.

Answer:

a. 104.5 C b. 7.94 × 10⁶ A/m² c. 5.83 × 10⁻⁴ m/s

Explanation:

a. How much electric charge passes through the starter motor?

Using Q = It where Q = electric charge passing through the starter motor, I = current = 110 A and t = time = 0.95 s

So, Q = It = 110 A × 0.95 s = 104.5 C

b. What is the current density in the wire?

The current density, J = I/A where I = current = 110 A and A = cross-sectional area = πd²/4 where d = diameter of copper wire = 4.20 mm = 4.20 × 10⁻³ m

So, J = I/A

= I/πd²/4

= 4I/πd²

= 4 × 110 A/π(4.20 × 10⁻³ m)²

= 440 A/55.42 × 10⁻⁶ m²

= 7.94 × 10⁶ A/m²

c. How far does an electron travel along the wire while the starter motor is on? (The density of conduction electrons in copper is n = 8.50×1028 1/m3.)

To find how far the electron travels, we need to find the electron drift velocity from

J = nev where J = current density = 7.94 × 10⁶ A/m², n = electron density = 8.50 × 10²⁸ m⁻³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of copper wire = 4.20 mm = 4.20 × 10⁻³ m

So, v = J/ne

Substituting the values of the variables into the equation, we have

v = 7.94 × 10⁶ A/m² ÷ (8.50 × 10²⁸ m⁻³ × 1.602 × 10⁻¹⁹ C)

v = 7.94 × 10⁶ A/m² ÷ (13.617 × 10⁹ Cm⁻³)

v = 0.583 × 10⁻³ m/s

v = 5.83 × 10⁻⁴ m/s

see the picture and here is the name of it

A = Pulmonary artery

B = inferior vena cava

C = aorta

D = left ventricle

E = right ventricle

F = pulmonary system in kidney

G = Pulmonary system

H = Pulmonary vein

#Moderators please don't be mean, dont delete my answers just to get approval from your senior or just to get the biggest moderation daily rank.

Answer:

Is fire matter? Matter is anything that has mass and occupies space. The flame itself is a mixture of gases (vaporized fuel, oxygen, carbon dioxide, carbon monoxide, water vapor, and many other things) and so is matter. The light produced by the flame is energy, not matter.

Answer:

The following seems to be the summary including its given phrase.

Explanation:

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

[tex]x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s[/tex]

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s  x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A  after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.