Ask question

Ask question

All categories

3 years ago

7

A 40kg kid riding their bike down the street at 5m/s reaches the edge of a hill and coast down to the bottom. If the hill was 10

m high, how much mechanical energy does the rider have at the bottom of the hill?

1 answer:

Romashka [77]3 years ago

6 0

Answer

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Explanation:v

You might be interested in

A circuit contains two resistors in series. The voltage drop across the first is 10 V. The voltage drop across the second is als

Norma-Jean [14]

Hi there!

Voltage in a series can be expressed by the following:

$V_T = V_1 + V_2$

In words, the total voltage is equal to the sum of the individual voltage drops in a SERIES circuit.

We can solve for the total voltage:

$V_T = 10 + 10 = \boxed{\text{ D. 20 V}}$

6 0

2 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago

un motor electric efectueaza un lucru mecanic de 864 j in 0.5 min.tensiunea la bornele sale este de 12v si este parcurs de un cu

GrogVix [38]

Answer:

96%

Explanation:

To find the values of the motor efficiency you use the following formula:

$E=\frac{P_o}{P_i}100$

P_o: output power = 864J/0.5min=864J/30s=28.8W

P_i: input power = I*V = (3A)(12V) = 36W

By replacing this values you obtain:

$E=\frac{28.8W}{30W}*100=96\%$

hence, the motor efficiency is about 96%

traslation:

Pentru a găsi valorile eficienței motorului, utilizați următoarea formulă:

P_o: putere de ieșire = 864J / 0.5min = 864J / 30s = 28.8W

P_i: putere de intrare = I * V = (3A) (12V) = 36W

Înlocuind aceste valori obțineți:

prin urmare, eficiența motorului este de aproximativ 96%

3 0

3 years ago

When bill butcher of port city brews discusses shelf space and the way that a merger could increase the leverage potential of la

lozanna [386]

He is discussing Bargaining Power of Buyers competitive force.

What do you understand by leverage?

The employment of various financial instruments or borrowed cash, or leverage, is an investing strategy that aims to improve an investment's potential return. The level of debt a company utilizes to finance its assets is another definition of leverage.

It offers a range of funding options so that the company can reach its desired earnings. Leverage is a crucial investing strategy because it enables businesses to establish a ceiling for the growth of their operations.

Leverage can be used, for instance, to support financially a new business. Purchasing fixed assets or borrowing money in the form of a loan from another business or person can serve as examples of leverage.

To learn more about leverage, visit: brainly.com/question/14230485

#SPJ4

6 0

2 years ago

Which is more reliable—using a manual stop watch or using light gates?<br> Explain.

Kitty [74]

Light gates are more reliable. When using a manual stop watch, it is difficult to stop it at an exact time. A light gate is able to detect when an object passes through a 'gate' with the infrared transmitter and receiver.

7 0

4 years ago