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Lilit [14]
3 years ago
13

Which of these equations will you used to find the final velocity if the initial

Physics
1 answer:
Svetach [21]3 years ago
5 0

Answer:

2.77 would be the answer for this

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Defenition of takes place in body cells ​
swat32

Answer:

the nucleus-containing central part of a neuron exclusive of its axons and dendrites that is the major structural element of the gray matter of the brain and spinal cord, the ganglia, and the retina. — called also perikaryon, soma.

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4 years ago
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Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en
Tom [10]

Explanation:

Wavelength in an emission spectrum,  \lambda=435\ nm=435\times 10^{-9}\ m  

The energy of an electron is given by :

E=\dfrac{hc}{\lambda}

Where

h is the Planck's constant

c is the speed of light

For 435 nm, the energy of the electron will be :

E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}

E=4.57\times 10^{-19}\ J

We know that 1\ eV=1.6\times 10^{-19}\ J

So, E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}

From the given energy levels :

E_5-E_2=-0.544-(-3.403)

E_5-E_2=2.859\ eV

So, the transition must be from E₅ to E₂. Hence, this is the required solution.                                             

5 0
4 years ago
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What provided the force that made the cart speed up
fgiga [73]

Answer:

Potential energy turn to kinetic energy

Explanation:

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4 years ago
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A ball is tossed straight up into the air with an initial speed of 12 m/s.
Nitella [24]

Answer:

a. ≈ 1.22449s

b. ≈ 14.69387m

c. ≈ 0.532415s

Explanation:

Because we are trying to find when it is at it's highest point we can safely say that it's velocity at that point is 0m/s,

therefore we can use the equation:

v_f = a(t_t) + v_i

and do some algebra to get:

\frac{v_f - v_i}{a} = t_t

Now we plug in our values (note that this is assumed to be on Earth and that because it says that upwards is positive, we set g to be negative to say that it is pointing down):

\frac{0- 12m/s}{-gm/s^2} = t_t

\frac{-12m/s}{-9.8m/s^2} = t_t

\frac{12}{9.8s} = t_t

1.22449s \approx t_t

To find the final height we can use:

x_f = \frac{v^2_f - v^2_i}{2a}

and plug in our values to get:

x_f = \frac{0^2m/s - 12^2m^2s^2}{-2g}

x_f = \frac{-144m^2s^2}{-2 \cdot 9.8m/s^2}

x_f = \frac{-144m^2s^2}{-19.6m/s^2}

x_f = \frac{144m}{19.6}

x_f \approx 14.69387m

To find the time we can use the time dependent position equation:

x_f = a(\frac{t^2_t}{2}) + v_i(t_t) + x_i

This here can be made into a quadratic equation like so (xi is set up to be 0m, so the equation wont have it):

at^2_t} + 2v_it_t - 2x_f = 0

Here we can use the quadratic formula:

t_t = \frac{-2v_i \pm \sqrt{(2v_i)^2 - 4(a)(-2x_f)} }{2(a)}

And now it would be best if we put in our values  (xf = 5m because that is our question):

t_t = \frac{-2(12m/s) \pm \sqrt{(2(12m/s))^2 - 4(-9.8m/s^2)(-2(5m))} }{2(-9.8m/s^2)}

t_t = \frac{(-24m/s) \pm \sqrt{(24^2m^2/s^2) + (39.2m/s^2)(-10m)} }{(-19.6m/s^2)}

t_t = \frac{(-24m/s) \pm \sqrt{(576m^2/s^2) + (-392m^2/s^2)} }{(-19.6m/s^2)}

t_t = \frac{(-24m/s) \pm \sqrt{(184m^2/s^2)} }{(-19.6m/s^2)}

t_t = \frac{(-24m/s) \pm (\sqrt{184}) m/s}{(-19.6m/s^2)}

Finally we have simplified enough to be worth solving for:

t_t = \frac{(-24) \pm (\sqrt{184})}{(-19.6s)}

We get:

t_t \approx 0.532415s \approx \frac{(-24) + (\sqrt{184})}{(-19.6s)}

and

t_t \approx 1.916564s \approx \frac{(-24) - (\sqrt{184})}{(-19.6s)}

Because time is always positive we want to choose the plus answer.

3 0
3 years ago
(correct answers only)
Gennadij [26K]

The time taken for the sailfish to reach the herring if it is 55.0 m away is 2.7 s.

<h3>Time of motion of the star fish</h3>

The time taken for the star fish to travel the given distance is calculated as follows;

s = ut + ¹/₂at²

55 = 11.8t + (0.5)(6.39)t²

55 = 11.8t + 3.195t²

3.195t² + 11.8t - 55 = 0

<em>solve the quadratic equation using formula method,</em>

where;

a = 3.195, b = 11.8, c = -55

t = 2.7 s

Thus, the time taken for the sailfish to reach the herring if it is 55.0 m away is 2.7 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

6 0
2 years ago
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