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Scrat [10]
3 years ago
9

WILL MARK BRAINLIEST!!

Physics
2 answers:
LenaWriter [7]3 years ago
6 0

Answer:

air above a wild fire

Explanation:

the type of current formed in the pot is referred to as a convectional current which forms when the heated fluid rises making the cool fluid descend the same happens with the air above a wildfire a convectional current will form when the warm air rises and the cool air descends.the main reason for tge descending and ascending is that the air or liquid closest to the source of heat will get heated making it lighter but the air or liquid tgat was further from the source of heat will not be as light so as the warm fluid rises, the cool fluid descends.

Rainbow [258]3 years ago
5 0

Answer:

air above a wildfire

Explanation

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It's a cloudy and rainy day. The air pressure is most likely _____.
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3 years ago
A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive dir
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Answer:

Magnetic field, B = 0.23 T          

Explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.

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3 years ago
Who was nikola teslas partner​
vesna_86 [32]

Answer:

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Explanation:

6 0
4 years ago
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A climber pulls down on a rope causing his body to rise up with the rope? Which law of motion is it?
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4 0
3 years ago
A simple pendulum has a period of 2.5 s. What is its period if its length is increased by a factor of four?
Svetach [21]

Answer:

Its period if its length is increased by a factor of four is 5 s.

Explanation:

The period of a simple pendulum is given by;

T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g}  \ be \ constant \\\\\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\

Given;

initial period, T₁ = 2.5

initial length, = L₁

new length, L₂ = 4L₁

the new period, T₂ = ?

\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\  T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\  T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s

Therefore, its period if its length is increased by a factor of four is 5 s.

5 0
3 years ago
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