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Serjik [45]
2 years ago
7

What fraction of a piece of concrete will be submerged when it floats in mercury? the density of concrete is 2.3×103kg/m3 and de

nsity of mercury is 13.6×103kg/m3?
Physics
1 answer:
chubhunter [2.5K]2 years ago
3 0
<span>17% Any object that is floating will display the volume of fluid that matches the mass of the object that's floating. So if the object is 100% as dense as the fluid, it will just barely sink, if it's 50% as dense as the fluid, 50% will be submerged, etc. So let's see what percent of mercury's density is concrete. 2.3x10^3 / 13.6x10^3 = 0.169117647 = 16.9117647% Rounding to 2 significant figures gives 17%. So 17% of a piece of concrete will be submerged in mercury when it's floating in mercury.</span>
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Ariel dropped a golf ball from her second story window. The ball starts from rest and hits the sidewalk 3.5 s later with a veloc
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Answer:

By using the acceleration formula,

a =  \frac{v - u}{t}

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2 years ago
A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th
agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

T - mg = ma

here we know that maximum possible acceleration of so that string will not break is given as

T = 54 N

now we have

54 - (5 \times 9.8) = 5 a

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d = \frac{1}{2}at^2

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7 0
2 years ago
Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation
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Following statements are true
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When the sweater and the leather jacket are in contact with each other, the leather jacket loses electrons and thus becomes positively charged. the electrons are gained by the sweater and it becomes negatively charges.
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A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant
OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

a_t = ar

where;

a is the angular acceleration and

r is the radius of the circular path

a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2

Determine time of the rotation;

\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2

The magnitude of total linear acceleration is given by;

a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27  \ m/s^2

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0
2 years ago
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