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Alja [10]
3 years ago
5

A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th

e water? b) What is the buoyancy force on the rock? c) What is the mass of the water that the rock displaces? d) What is the weight of the rock under water?
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

   Volume = 470 cm³

   Density = 3.55 g/cm³

   Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

   Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

   Volume = 470 cm³

   Density of liquid = 1 g/cm³

   \texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N

c) Mass of the water displaced = Volume of body x Density of liquid

   Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

   Weight of rock under water = 16.37 - 4.61  =11.76 N

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Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
The measurement 0.025563 g should be reported as?
Tomtit [17]
The given mass is 0.025563 g.

Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.

b. 2.5 x 10² g = 250 g.
It is incorrect.

c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.

d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.

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7 0
3 years ago
The bob of a pendulum swings back and forth with a total mechanical energy of 300 J. What is the kinetic energy of the bob when
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at the lowest point in the trajector, the kinetic energy of the bob is 300 J.

Explanation:

The total mechanical energy of the bob at any point of its motion is given by

E=KE+PE

Where

KE=\frac{1}{2}mv^2 is the kinetic energy, where

m is the mass of the bob

v is its speed

PE=mgh is the gravitational potential energy, where

g is the acceleration of gravity

h is the height of the bob, measured with respect to the lowest point of the trajector

In absence of friction, the total mechanical energy E remains constant. So we have:

- When the bob swings upward, the PE increases (because h increases) and the KE decreases (so the speed decreases). At the highest point in the trajector, the speed of the bob is zero (v=0), so its KE is also zero and all the mechanical energy is potential energy: U = 300 J

- When the bob swings downward, the PE decreases (because h decreases) and the KE increases (so the speed increases). At the lowest point in the trajectory, the height has become zero (h=0), so the PE is zero and all the mechanical energy is kinetic energy: KE = 300 J

Therefore, at the lowest point in the trajector, the kinetic energy of the bob is 300 J.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

4 0
3 years ago
Mass m, moving at speed 2v, approaches mass 4m, moving at speed v. The two collide elastically head-on. Part A Find the subseque
meriva
Thank you for posting your question here at brainly. A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V. I hope it helps.
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A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured t
kolbaska11 [484]

The material that the cylinder is made from is Butyl Rubber.

<h3>What is Young's modulus?</h3>

Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

<h3>Area of the cylinder</h3>

A = πr²

A = \pi \times (0.02)^2 = 0.00126 \ m^2

<h3>Young's modulus of the cylinder</h3>

E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae} \\\\

Where;

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When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

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When the mass is 50 kg,

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The Young's modulus is between 0.001 GPa  to 0.002 GPa

Thus, the material that the cylinder is made from is Butyl Rubber.

Learn more about Young's modulus here: brainly.com/question/6864866

3 0
3 years ago
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