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2 years ago

12

"On a good day, it takes Mr. Hess 23 minutes (1380 seconds) to drive the 12 miles (19,300 meters). What is his average velocity

on the way to school?"
I uh, don't understand this at all. If someone could help me solve this, I'd greatly appreciate it!

1 answer:

il63 [147K]2 years ago

5 0

Answer:

his average velocity is

$v = \frac{19300}{1380} = 13.9855(m \div s)$

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Sometimes, supplies can be delivered to people in remote areas by dropping them off from a plane. This is usually the case for g

FrozenT [24]

The order the boxes would land from slowest to quickest is A,C,B which is option C.

<h3>What is a Parachute?</h3>

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The size of the parachute affects the speed of falling because a larger parachute allows it to displace more air, causing it to fall more slowly. The one with three parachutes which is A will fall slowest and the one with just one will fall fastest.

Read more about Parachutes here brainly.com/question/499768

8 0

2 years ago

If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Tasya [4]

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

7 0

3 years ago

Sometimes balance point may not be obtained on the potentiometer wire why

scoundrel [369]

Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

.
The balance point is not on the potentiometer wire - this statement means that

. In that case ,
l > L
V > E

8 0

2 years ago

A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670

Nikolay [14]

Answer: 459.14 N

Explanation:

from the question, we have

diameter = 10 m

radius (r) = 5 m

weight (Fw) = 670 N

time (t) = 8 seconds

Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below

∑ F = F c = F w − Fn ..............equation 1

Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2

substituting the value of v as (2πr / T) we now have

Fn = mg − (m(2πr / T )^2) / r

Fn= mg − (4(π^2)mr / T^2) ..........equation 3

Fw (mass of the person) = mg

therefore m = Fw / g

m = 670 / 9.8 = 68.367 kg

now substituting our values into equation 3

Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)

Fn = 670 - 210.86

Fn = 459.14 N

4 0

3 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for

skelet666 [1.2K]

Answer:

λ = 162 10⁻⁷ m

Explanation:

Bohr's model for the hydrogen atom gives energy by the equation

$E_{n}$ = - k²e² / 2m (1 / n²)

Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer

The Planck equation

E = h f

The speed of light is

c = λ f

E = h c /λ

For a transition between two states we have

$E_{n}$ - $E_{m}$ = - k²e² / 2m (1 / $n_{f}$ ² -1 / $n_{i}$ ²)

h c / λ = -k² e² / 2m (1 / $n_{f}$ ² - 1/ $n_{i}$ ²)

1 / λ = (- k² e² / 2m h c) (1 / $n_{f}$ ² - 1/ $n_{i}$ ²)

The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses

Let's calculate the emission of the transition

1 /λ = 1.097 10⁷ (1/10² - 1/8²)

1 / λ = 1.097 10⁷ (0.01 - 0.015625)

1 /λ = 0.006170625 10⁷

λ = 162 10⁻⁷ m

3 0

3 years ago