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Alex777 [14]
3 years ago
9

A ball is dropped from a height of 10 meters how long was the ball in the air and what is the speed when it hits the ground

Physics
1 answer:
Burka [1]3 years ago
6 0
4 Because it will go fast
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What amount of energy is needed for an electron to jump from n = 1 to n = 4?
liberstina [14]

Answer:

E=2.04\times 10^{-18}\ J

Explanation:

We need to find the energy for an electron to jump from n = 1 to n = 4.

The energy in transition from 1 state to another is given by :

E=\dfrac{-2.18\times 10^{-18}}{n^2}\ J

The difference in energy for n = 1 to n = 4 is:

E=-2.18\times 10^{-18}\times (\dfrac{1}{4^2}-1)\\\\E=2.04\times 10^{-18}\ J

So, the required energy is equal to 2.04\times 10^{-18}\ J.

4 0
3 years ago
someone made a copy of an analog wave in a digital wave compare the copies with the original waves and explain which type of wav
lilavasa [31]

Answer: Analog transfer information quicker, example electromagnetic waves which are used in our modern day phones.

Explanation:  Digital signals are a more reliable form of transmitting information because an error in the amplitude or frequency value would have to be very large in order to cause a jump to a different value. Signals are composed of infinite possible values. Signals are composed of only two possible values: 0 or 1.

6 0
3 years ago
Read 2 more answers
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0
3 years ago
What is the combined resistance for two resistors with resistances of 10 Ohms and 23.7 Ohms if they are connected in series
Agata [3.3K]

Answer: 33.7Ω

Explanation:

Since there are two resistors connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2

R1 = 10Ω

R2 = 23.7Ω

Hence, Rtotal = 10Ω + 23.7Ω

Rtotal = 33.7Ω

Thus, the combined resistance for two resistors is 33.7Ω

3 0
3 years ago
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