Answer:
The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel
" applies to electric charges.
Explanation:
There are only two types of electric charges. Both having own magnitude but different charge.
1. Positive charge
2. Negative charge
Like charges repel each other and opposite charges always attract each other.
When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.
Answer:
r₂ = 0.2 m
Explanation:
given,
distance = 20 m
sound of average whisper = 30 dB
distance moved closer = ?
new frequency = 80 dB
using formula

I₀ = 10⁻¹² W/m²
now,



to hear the whisper sound = 80 dB



we know intensity of sound is inversely proportional to square of distances



r₂ = 0.2 m
When a person has experienced a tire blow out, there are steps to follow in order to prevent harm to others and self. After following the required and helpful steps after a tire blow out, slowed down and regain control, it is best that the person should head to a stop road or at the safe side of the lane where they won't be a bother to the road or to people driving in the high way. After pulling over to the side, it is advisable to turn on the emergency flashers of the car. This will set as a signal that you are in need of help or had gone through a problem.
Answer:
(I). The resistance of the copper wire is 0.0742 Ω.
(II). The resistance of the carbon piece is 1.75 Ω.
Explanation:
Given that,
Length of copper wire = 1.70 m
Diameter = 0.700 mm
Length of carbon piece = 20.0 cm
Cross section area
(I). We need to calculate the area of copper wire
Using formula of area


We need to calculate the resistance
Using formula of resistance

Put the value into the formula


(II). We need to calculate the resistance
Using formula of resistance

Put the value into the formula


Hence, (I). The resistance of the copper wire is 0.0742 Ω.
(II). The resistance of the carbon piece is 1.75 Ω.
Answer:
an inhibitor of angiotensin II
Explanation:
Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.