Gamma radiation could be used to sterilize plastic petri plates in a plastic wrapper.
Answer: Option E
<u>Explanation:
</u>
It is known that radioactive radiations have greater ionizing powers compared to other electromagnetic radiations. Among the radioactive radiations, gamma rays are the most ionizing radiation.
So they are widely used in sterilizing process. As the penetration power of gamma is more compared to alpha and beta, in order to kill microorganisms gamma radiations are used.
When plastic petri plates wrapped with plastic are bombarded or sterilizing with gamma rays, the rays will ionise the micro-organism. Thus mutilating their DNA and preventing their reproduction.
EC_1 + EP_1 = EC2 + EP_2
EC_2 = 0
EC_2 = EP_1 - EP_2
EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
Answer:
c. is more than that of the fluid.
Explanation:
This problem is based on the conservation of energy and the concept of thermal equilibrium
m= mass
s= specific heat
\DeltaT=change in temperature
let s1= specific heat of solid and s2= specific heat of liquid
then
Heat lost by solid= 
Heat gained by fluid=
Now heat gained = heat lost
therefore,
1000 S_2=800 S_1
S_1=1.25 S_2
so the specific heat of solid is more than that of the fluid.
Answer: 20.4752789138x x 10^23 atoms
To count how many atoms in moles you need to know Avogadro's number. Avogadro's number dictate that for every mole there is 6.022140857 × 10^23 molecule/atoms.
Then 3.4 moles of helium will be 3.4x 6.022140857 x 10^23 atoms= 20.4752789138x x 10^23 atoms
Answer:
the money that would be saved is $13.14.
Explanation:
Given;
power consumed by the light bulb, P = 100 W = 0.1 kW
time of running the bulb, t = 3 hours for 365 days = 1,095 hours
cost rate of power consumption, C = $0.12 per kWh
Energy consumed by the light bulb for the given days;
E = Pt
E = 0.1 kW x 1,095 hr
E = 109.5 kWh
Cost of energy consumed = 109.5 kWh x $0.12 / kWh
= $13.14
Therefore, the money that would be saved is $13.14.
