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Elza [17]
3 years ago
8

The game of kickball begins with the ball being rolled toward a player. The player then kicks the ball across the field. What is

true about the speed of the ball when it is kicked?
A.
The speed of the ball is not dependent on the force used to kick it.

B.
The speed of the ball will be slower than the speed at which it was pitched.

C.
The speed of the ball depends on the force used to kick it.

D.
The speed of the ball mostly depends on the speed at which it was pitched.
which is correct
Chemistry
2 answers:
hammer [34]3 years ago
6 0

Answer:

C. The Speed of the ball depends on the force used to kick it.

Explanation:

Olegator [25]3 years ago
4 0

Answer:

d is correct

Explanation:

a ball loses some force of the pitch when touching a solid object and so

D.

The speed of the ball mostly depends on the speed at which it was pitched.

which is correct

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musickatia [10]

Answer:

520.12554

Explanation:

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2 years ago
What is the wavelength of light that has a frequency of 3.7 x 1014 Hz? Show all work!!!
kramer

Answer:

Explanation:

f=\frac{c}{\lambda}=>\lambda=\frac{c}{f}=\frac{300x10^8\frac{m}{s}}{3.7*10^{14}\frac{1}{s}}=810.8108*10^{-9}m

3 0
3 years ago
Read 2 more answers
Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

5 0
3 years ago
Which of the following pairs of reactants is likely to produce a pure metal when the
storchak [24]
1. Na3PO4+Mg(NO3)2= NaNO3 + MgPO4
So this is not the answer.
2. Al+CuCl2= AlCl3+ Cu
So this is the answer.
3. Mg+O2= MgO2
So this is not an answer
4. Cl2+KI= KCl+I2
So this is also an answer

Hence B and C are answers
3 0
3 years ago
An aqueous solution of barium hydroxide is standardized by titration with a 0.102 M solution of perchloric acid. If 10.3 mL of b
Vesna [10]

<u>Answer:</u> The molarity of barium hydroxide solution is 0.118 M.

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HClO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ba(OH)_2

We are given:

n_1=1\\M_1=0.102M\\V_1=24.0mL\\n_2=2\\M_2=?M\\V_2=10.3mL

Putting values in above equation, we get:

1\times 0.102\times 24.0=2\times M_2\times 10.3\\\\M_2=0.118M

Hence, the molarity of Ba(OH)_2 solution will be 0.118 M.

6 0
3 years ago
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