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Nataly_w [17]
4 years ago
15

What are the equations used to describe work?

Physics
1 answer:
Phantasy [73]4 years ago
8 0
Work = force x displacement x cosine(theta)

easier way: 
W = F x d x cos(theta)
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How many photons are absorbed during the dental x-ray?
MariettaO [177]
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.

Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
8 0
3 years ago
If a car is travelling 120 miles southbound for 3 hours, what is the velocity?
Zinaida [17]

Answer:

40 miles / hour south

Explanation:

120 miles/3 hours = 40 miles / hour

4 0
3 years ago
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What is the definition of permitted orbits ?
givi [52]
.The path of a celestial body or an artificial satellite as it revolves around another body due to their mutual gravitational <span>attraction.</span>
7 0
4 years ago
A helicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. de
uranmaximum [27]
Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.

Let a =  the acceleration of descent., m/s².
Let u =  initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u,  v, and a are measured as positive upward.

Then
 u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a                  (1)

Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600       (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²

From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s

Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.

3 0
3 years ago
How long would it take the Earth to complete 4 orbits around the Sun?
yuradex [85]

Each orbit is approximately 365.24 days ... the period we call a "year".

Four of those = 1,461 days  =  4 years .

6 0
4 years ago
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