Answer:
The Kinetic Energy is approximately 3 times decreased
Explanation:
A baseball weighs 5.13 oz.
a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?
b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.
Kinetic Energy (KE)=0.5×mass×velocity ^ 2
Kinetic Energy (KE)=0.5×mass × velocity ^ 2
Joules = kg×m^2/s^2
1 mile = 1609.344 meters
1 hour = 3600 sec
1 Oz = 28.34952 g = 0.02834952 kg
a) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=130.761 kg×m^2/s^2 = 130.761 Joules
b) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=43.51028 kg×m^2/s^2 = 43.51028 Joules
= 130.761 / 43.51028 = 3.00528,
As such the Kinetic Energy is approximately 3 times decreased
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
The equation that scientists could use to find the wavelength of the emission lines of the hydrogen atom would be that of Balmer.
The wavelength of the emission lines of the hydrogen atom can be derived using the Balmer series:
1/λ 
Where λ = wavelength, 
= Rydberg constant, and n = level of the original orbital.
The equation becomes applicable in getting the wavelength of emitted light when electrons in hydrogen atoms transition from higher (n) orbital to lower orbital (2) levels.
More on the Balmer series can be found here: brainly.com/question/5295294
