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pav-90 [236]
2 years ago
9

Which diagram shows how Rachel can see a candle flame?

Physics
1 answer:
SIZIF [17.4K]2 years ago
5 0

Answer:

C

Explanation:

If the arrows represent light rays, then Rachel sees a candle flame when the light released by the flame is received by her eyes.

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Hurricane katrina, which hit the gulf coast of louisiana and mississippi on august 29, 2005, had the second lowest ever recorded
Simora [160]
The unit 'mb' means millibar which is equivalent to 1/1000 of 1 bar. To convert the units from bar to atmospheres (atm) and to inches Hg (inHg), we need to know the conversion factors.

a.) 1 atm = 1.01325 bar

0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) =<em> 9.08×10⁻⁴ atm</em>

b.) 1 bar = 29.53 inHg

0.92 mb(1 bar/1000 mbar)(29.53 inHg/1 bar) =<em> 0.027 inHg</em>
3 0
2 years ago
A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
The environment affects people, but people don't affect the environment.<br><br><br>TRUE OR FALSE
Drupady [299]

Answer: False

Explanation:

The statement that the environment affects people, but people don't affect the environment is incorrect. It should be noted that people also affect the environment.

Some of our impact on the environment include pollution, deforestation, overpopulation, burning of fossil fuels etc. Thus has resulted in changes such as erosion, climate change, etc.

5 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the
kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0
2 years ago
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