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3 years ago

If element X has 96 protons, how many electrons does it have

Physics

1 answer:

NNADVOKAT [17]3 years ago

3 0

It would have 96 because protons equal electrons

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A pure substance can be a ..................... (a) element (b) compound (c) either element or compound (d)none of these

stiks02 [169]

Answer:

Explanation:

An element is a pure substance that can not be broken down into anything simpler

A compound in also a pure substance held together in fixed proportion through chemical bonds

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3 years ago

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A baseball is located at the surface of the earth. Which statements about it are correct? (There may be more than one correct ch

lakkis [162]

i think that the answer is a)

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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

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3 years ago

How far from the castle wall does the launched rock hit the ground?

lina2011 [118]

King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.

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3 years ago

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A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a

mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required $t=\frac{distance}{speed}=\frac{d}{38}hour$

Speed when car return from hill = 66 km/h

So time required to return fro hill $t=\frac{d}{66}h$

Total time $t_{total}=\frac{t}{38}+\frac{t}{66}$

Total distance = d+d =2d

So average speed $=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h$

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3 years ago