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ch4aika [34]
3 years ago
6

Can someone help me??? PLZ. ASAP. TODAY!!!!! Only number 1#1

Physics
1 answer:
il63 [147K]3 years ago
4 0
The answer is letter a
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Which statement describes the relationship of resistance and current?
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Choices 'B'; and 'D' both begin with the correct words.
But they should end with the equation

                       R = V / I 
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Undersea mountain ranges in the middle of the ocean floors are known as d͟e͟e͟p͟-͟o͟c͟e͟a͟n͟ ͟t͟r͟e͟n͟c͟h͟e͟s.
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mid-ocean ridge

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What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?
Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, q=6.4\times 10^{-6}\ C

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV

So, the required electric potential is 192 kV.

3 0
3 years ago
A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu
satela [25.4K]
<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>
3 0
3 years ago
A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
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