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tester [92]
3 years ago
8

1. What is the diameter of Mars?

Physics
1 answer:
Katen [24]3 years ago
3 0
  1. 6,799 km
  2. On average, the data revealed, the equatorial Martian atmosphere consists of 95% carbon dioxide, 2.59% nitrogen, 1.94% argon, 0.161% oxygen, and 0.058% carbon monoxide.

3. The simple explanation for the Red Planet's color is that its regolith, or surface material, contains lots of iron oxide — the same compound that gives blood and rust their hue. ... It does have an iron core, but abundant iron exists in its upper layers, also.

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An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
4 years ago
Second and third class levers are differentiated by __________. a. the location of the fulcrum b. the location of the load c. th
strojnjashka [21]

Answer:

the answer is b

Explanation:

Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>

<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>

5 0
2 years ago
A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released fr
poizon [28]

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t

5 0
3 years ago
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