Answer:
17 kJ
Explanation:
Calculation for the Calculate the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C.
Using this formula
q = mC∆T
Where,
q represent Energy
m represent Mass of substance=0.60kg=600g
C represent Specific heat capacity=2.44J·g−1K−1.
∆T represent change in Temperature=2.2°C to 13.7°C.
Let plug in the formula
q=(0.60 kg x 1000 g/kg)(2.44 J/gº)(13.7°C-2.2°C)
q = (600g)(2.44 J/gº)(11.5º)
q=16.836 kJ
q= 17 kJ (Approximately)
Therefore the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C will be 17 kJ
The answer for the problem is explained below.
The option for the answer is "D".
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Explanation:
Given:
wavelength (λ) = 468 nm = 468×10^-9 m
speed of light (c) = 3.00 x 10^8m/s
Planck's constant is 6.626 x 10^-34J·s
To solve:
energy of light (E)
We know,
E =(h×c) ÷ λ
E = ( 6.626 x 10^-34 × 3.00 x 10^8) ÷ 468×10^-9
E = 4.25 × 10^-19 J
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Answer:
Radiation effects on electrical equipment depend on the equipment and on the type of ionizing radiation to which it is exposed.
First, beta radiation has little, if any, effect on electrical equipment because this type of ionizing radiation is easily shielded. The equipment housing and the construction of the parts within the housing will protect the equipment from beta-radiation (high-energy electrons) exposure.
Gamma radiation is penetrating and can affect most electrical equipment. Simple equipment (like motors, switches, incandescent lights, wiring, and solenoids) is very radiation resistant and may never show any radiation effects, even after a very large radiation exposure. Diodes and computer chips (electronics) are much more sensitive to gamma radiation. To give you a comparison of effects, it takes a radiation dose of about 5 Sv to cause death to most people. Diodes and computer chips will show very little functional detriment up to about 50 to 100 Sv. Also, some electronics can be "hardened" (made to be not affected as much by larger gamma radiation doses) by providing shielding or by selecting radiation-resistant materials.
Some electronics do exhibit a recovery after being exposed to gamma radiation, after the radiation is stopped. But the recovery is hardly ever back to 100% functionality. Also, if the electronics are exposed to gamma radiation while unpowered, the gamma radiation effects are less.
Ionizing radiation breaks down the materials within the electrical equipment. For example, when wiring is exposed to gamma rays, no change is noticed until the wiring is flexed or bent. The wire's insulation becomes brittle and will break and may cause shorts in the equipment. The effect on diodes and computer chips is a bit more complex. The gamma rays disrupt the crystalline nature of the inside of the electronic component. Its function is degraded and then fails as more gamma radiation exposure is received by the electronic component.
Gamma rays do not affect the signals within the device or the signals received by the device. Nonionizing radiation (like radio signals, microwaves, and electromagnetic pulses) DO mess with the signals within and received by the device. I put a cheap electronic game in my microwave oven at home. It arced and sparked and was totally ruined. I didn’t waste any more of my time playing that game.
Hope this helps.
Explanation:
MARK ME AS BARINIEST PLS
If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.
If the angle is 45, then there is still some work involved.
The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90.
