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Arte-miy333 [17]
3 years ago
9

the two forces acting on a falling object are gravity and?....1.force, 2. friction, 3. air resistance, and 4. net force?​

Physics
1 answer:
Vsevolod [243]3 years ago
6 0
In free fall, the only force acting on an object is gravity. The force of gravity is an unbalanced force, which causes an object to accelerate. In the absence of air, two objects with different masses fall at exactly the same rate of acceleration.
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what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three
grin007 [14]

Answer:

Approximately 5.11 \times 10^{-19}\; {\rm J}.

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}.

Look up the speed of light in vacuum: c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}.

Look up Planck's constant: h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}.

Apply the Rydberg formula to find the wavelength \lambda (in vacuum) of the photon in question:

\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}.

The frequency of that photon would be:

\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}.

Combine this expression with the Rydberg formula to find the frequency of this photon:

\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}.

Apply the Einstein-Planck equation to find the energy of this photon:

\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}.

(Rounded to three significant figures.)

6 0
2 years ago
Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi
Leviafan [203]

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

<u>W = 12.96 J</u>

5 0
3 years ago
How does the number of valence electrons affect the reactivity?
ololo11 [35]
The less valence electrons the more reactive the element is.
6 0
3 years ago
A pendulum bob is given velocity u =
Mama L [17]

Answer:

V = P = 0 m/s

Explanation:

When a pendulum bob is given an initial displacement or the initial velocity, it starts to execute periodic motion or simple harmonic motion. During this motion the kinetic and potential energy keeps interconverting. The kinetic energy becomes maximum at the lowest point, that is the mean point. Hence, the velocity is maximum at this point, as well. Similarly, at both extreme positions the potential energy becomes maximum due to maximum height, while the kinetic energy becomes zero at the highest point, that is extreme positions. At these, positions the velocity will be minimum and it will be zero due to zero kinetic energy. Hence, at both extreme positions the bob stops momentarily before, reversing the direction. Hence,

<u>V = P = 0 m/s</u>

7 0
3 years ago
Two forces F1 and F2 are acting on a block of mass m=1.5 kg. The magnitude of
nika2105 [10]

Answer:

Explanation:

Component of force F₁ in right direction = F₁cos37

= 12 x cos37 = 9.58 N .

Component of force F₂ in  vertically upward direction = F₁sin37

= 12 x sin37 = 7.22 N .

a ) Let normal force be R

R + F₁sin37  = mg

R + 7.22 = 1.5 x 9.8 = 14.7

R = 7.48 N .

b )

Net force in horizontal direction = F₂ - F₁cos37

= F₂ - 9.58

This is equal to zero as body is moving with zero acceleration

F₂ - 9.58 = 0

F₂ = 9.58 N

c ) If body is moving with acceleration of 2.5 m /s² along the direction of F₂

F₂ - 9.58 = 1.5 x 2.5 = 3.75

F₂ = 13.33 N .

6 0
3 years ago
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