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____ [38]
3 years ago
14

The mass density of an object is 16.3 g/cc, and its volume is determined to be 16.7 cc. what is the mass of the object?

Physics
1 answer:
guajiro [1.7K]3 years ago
3 0
There are 16.3 grams for every cubic centimeter, and there are 16.7 cubic centimeters. therefore,
16.3 x 16.7

=272.21 g
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Rashid [163]

Answer:

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3 years ago
A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the
LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

\rho_w = \frac{m_w}{V_w}

If we solve for mw we got m_w = \rho_w V_w  (2)

Replacing equation (2) into equation (1) we got:

B = \rho_w V_w g (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0
3 years ago
Help pls lol and tysm
Zanzabum

Answer:

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Explanation:

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7 0
3 years ago
What type of wave vibrates parallel to the direction of travel
Irina-Kira [14]

Answer: Longitudinal waves

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5 0
3 years ago
A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered 20 m? What would be its velocity w
stellarik [79]

Given :

Height from which ball is dropped , h = 40 m .

Acceleration due to gravity , g= 10 m/s² .

Initial velocity , u = 0 m/s .

To Find :

Velocity when ball covered 20 m and velocity when it hit the ground .

Solution :

Now , height when ball covered 20 m distance is , 40 - 20 = 20 m .

By equation of motion :

v^2=u^2+2gh\\\\v=\sqrt{2\times 10\times 20}\ m/s\\\\v=20\ m/s

Now , distance covered when body reaches ground is , 40 m .

Putting value h = 40 m in above equation , we get :

v=\sqrt{2\times 10\times 40}\ m/s\\\\v=20\sqrt{2}\ m/s

Hence , this is the required solution .

7 0
3 years ago
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