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Damm [24]
3 years ago
8

A rock falls off a cliff and hits the ground after three seconds. What is its velocity right before it hits the ground?

Chemistry
2 answers:
mezya [45]3 years ago
5 0

The correct answer is 29.4 m/s.

The right formula to use for the scenario given in the question is:

Vf = Vi + at

Where: Vf = Final velocity = ?

Vi = Initial velocity = 0

a = Acceleration [due to gravity] = 9.8 m/s

T = Time = 3 s

Note that an object that is falling down from a height has an initial velocity of zero.

Vf = 0 + [9.8 * 3] = 29.4

Therefore, final velocity = 29.4 m/s.

Amiraneli [1.4K]3 years ago
4 0
To solve this kinematics formula use the following equation:

Vf = Vi + at
Vf = 0 + (9.81 m/s^2)(3 seconds)
Vf = 29.43 m/s and or about 29.4 m/s of reported to 3 significant figures.
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5.Principal, Azimuthal (subsidiary), and magnetic quantum number are the first three because 2 stands for principal, s-for azimuthal (l=0) and magnetic quantum number for s- orbital= 0

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3 years ago
What is the pH of a 0.0042 M hydrochloric acid solution?
babymother [125]

Answer:

E) 2.38

Explanation:

The pH of any solution , helps to determine the acidic strength of the solution ,

i.e. ,

  • Lower the value of pH , higher is its acidic strength

and ,

  • Higher the value of pH , lower is its acidic strength .

pH is given as the negative log of the concentration of H⁺ ions ,

hence ,

pH = - log H⁺

From the question ,

the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,

Therefore , the concentration of H⁺ = 0.0042 M .

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3 years ago
Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

 T_c=1181.55k

Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

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