Given the distance from the through of one wave to the next wave you can determine the HELP FAST!!!!!

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

Nesterboy [21]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

7 0

3 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

3 years ago

A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the roc

vovangra [49]

Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

$v_x = 2.70 m/s$

Part b)

In vertical direction we can use kinematics

$v_f = v_i + at$

$v_f = 0 + 2.1 * 9.8$

$v_f = 20.6 m/s$

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

5 0

2 years ago

According to Newton’s Third Law, action and reaction are

Komok [63]

<span>B. equal and in opposite directions</span>

5 0

3 years ago

Read 2 more answers

2 ways in which a weather satellite can orbit the earth

avanturin [10]

Answer:

The two most common types of orbit are "geostationary" and "polar."

6 0

2 years ago

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