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
Coins or anything copper
Explanation
They are everyday objects that turn green.
Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
There will not be any ejection of photoelectrons
Explanation:
Energy of the photon= hc/λ
Where;
h= Plank's constant
c= speed of light
λ= wavelength of the incident photon
E= 6.6×10^-34 × 3 ×10^8/488 × 10^-9
E= 4.1 ×10^-19 J
Work function of the metal (Wo)= 2.9 eV × 1.6 × 10^-19 = 4.64 × 10^-19 J
There can only be ejected photoelectrons when E>Wo but in this case, E<Wo hence there will not be any ejection of photoelectrons.
