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3 years ago

A large object las Less of an attraction on a smaller True False

1 answer:

6 0

Do you mean ‘Does a larger object have less attraction than a smaller object?’ If so, then the answer is objects with a larger mass exert more attraction while objects with a smaller mass have less attraction.

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Which form of energy body will possess when it is kept on the top of Hill

Oduvanchick [21]

Potential energy would be the answer

3 0

3 years ago

Read 2 more answers

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

On a hot day, the deck of a small ship reaches a temperature of 48

AlekseyPX

The final temperature of the seawater-deck system is 990°C.

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

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6 0

2 years ago

8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro

omeli [17]

The distance that the canoe moves in this process is 1.29 meters.