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4 years ago

14

A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

water? The density of lead is 11.4 g/cm³.

1 answer:

Kruka [31]4 years ago

4 0

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

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Answer:

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3 years ago

Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and

monitta

Answer:

$a=2,5m/s^2$

Explanation:

From the question we are told that:

Coefficient of kinetic friction $\mu= 0.200$

Vertical Mass $M_v=3kg$

Horizontal mass $M_h=3.00kg$

Generally the equation for kinetic force $F_k$ is mathematically given by

$F_k=\mu N\\F_k=0.2*3\\F_k=0.6$

Generally the equation for T is mathematically given by

$For M_v=3kg3g-T=3a$

For $M_h=3kg$

$T=M_v V+F_k\\T=3.0a+0.6$

Therefore substituting

$3-3a-0.6=3a\\2.4g=6a$

$a=2,5m/s^2$

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3 years ago

A model rocket fired vertically from the ground ascends with a constant vertical acceleration of 52.7 m/s2 for 1.41 s. its fuel

olya-2409 [2.1K]

For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is

a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.

-52.7 = (0-v1)/1.41
v1 = 74.31 m/s

We can use v1 for the formula of the maximum height attained by an object thrown upwards:

Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m

The maximum height attained by the model rocket is 281.42 m.

For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.

Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.

2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)

Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s

Therefore, the total time= 1.41+6.83+7.57 = 15.81 s

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3 years ago

Assume that two cars have the same kinetic energy, but that the red car has twice the speed of the blue car. We then know that t

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Answer:

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3 years ago

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

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The ball's horizontal and vertical velocities at time $t$ are

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$y=20.0\,\mathrm m-\dfrac g2t^2$

The ball travels 22 m horizontally from where it was thrown, so

$22\,\mathrm m=v_{xi}t$

from which we find the time it takes for the ball to land on the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it lands, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

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3 years ago