What height would a 4 kg book need to be to have a potential energy of

Now put that book 5 feet in the air. What kind of energy does the book have? Explain.

erastova [34]

Answer:

Gravitational potential energy

Explanation:

The book is put 5 feet in the air, which means 5 feet above the ground. An object which is located to a certain height above the ground possesses a form of energy called gravitational potential energy, which is the energy due to the fact that the object has "potential" to transform this energy into other forms of energy (e.g. kinetic energy, if the book is released and it starts moving).

The value of the gravitational potential energy of the object is given by the formula:

$U=mgh$

where

m is the mass of the object

g is the gravitational acceleration (9.8 m/s^2)

h is the height of the book above the ground (in this case, 5 feet)

So, we see that the gravitational potential energy is proportional to both the mass and the height of the object.

8 0

4 years ago

The directional wave patterns or illusion of motion in the hair are known as the:________

disa [49]

The answer is Design texture.

Hair with directional wave patterns or motion illusions. When designing a style, design texture must be considered. Curly hair reflects less light and has a greater shape than straight or wavy hair.

What is Design texture?

Texture is a design feature that delineates the surfaces of shapes and forms.
Tactile texture is texture that you feel with your fingertips, whereas visual texture is texture that an artist recreates on a flat surface.
Because it possesses height, breadth, and depth, tactile texture is three-dimensional.

To learn more about Design texture visit:

brainly.com/question/14832382

#SPJ4

5 0

2 years ago

Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel

kenny6666 [7]

Explanation:

Given that,

Two resistors 4.5 Ω and 2.3 Ω .

Potential difference = 30 V

When they are in series, the current through each resistor remains the same. First find the equivalent resistance.

R' = 4.5 + 2.3

= 6.8 Ω

Current,

$I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A$

So, the current through both lightbulb is the same i.e. 4.41 A.

When they are in parallel, the current divides.

Current flowing in 4.5 resistor,

$I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A$

Current flowing in 2.3 ohm resistor,

$I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04$

In parallel combination, are brighter than bulbs in series.

5 0

3 years ago

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

What is the total kinetic energy and potential energy of a object or group of objects

expeople1 [14]

it is a part of gravity because it brings you back down

7 0

3 years ago