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olasank [31]
2 years ago
10

The side length of a square is modeled by s=A, where A is the area of the square. Graph the function. What is the side length of

a square with an area of 35 square units?
Physics
1 answer:
Naya [18.7K]2 years ago
3 0

Answer:

√35 square units

Explanation:

Area of square = 35square units

Formula of area of square is a =s*s where a is area and s is side

So,

s*s = 35

s^2= 35

s = √35

tell me if this is the answer u r looking for

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What rapid changes to earths surface are caused by shifting plates
PtichkaEL [24]
I am not sure if this is the answer you are looking for but a earthquake occurs when the plates shift. 
Hope this helped. 
4 0
3 years ago
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).
qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

or

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

 

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

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
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a
Troyanec [42]

Answer: 6.47m/s

Explanation:

The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

V = 2π * 0.309 / 0.3

V = 1.94/0.3

V = 6.47m/s

The tangential speed of the tack is 6.47m/s

7 0
3 years ago
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If the gears are of different sizes, they can be used to increase the power of a turning force. The smaller wheel turns more quickly but with less force, while the bigger one turns more slowly with more force. Cars and bicycles use gears to achieve amazing speeds our bodies could never match without help.

3 0
3 years ago
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