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storchak [24]
3 years ago
13

The element lead (Pb) has a density 11.3 times that of water. Copper (Cu) has a density 7.9 times the density of water. A 5 kg m

ass of lead and a 5 kg mass of copper are both completely submerged in a bucket of water. Which mass has the LARGER buoyant force acting on it?
A) The buoyant force on the lead mass is larger.
B) The buoyant force on the copper mass is larger.
C) The buoyant force is the same on both masses.
Physics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

The answer is B

Explanation:

Density of the element lead (Pb) is:

d_{Pb} =11,3kg/dm^3

Density of the element Copper (Cu) is:

d_{Cu} =7,9kg/dm^3

First we need o find the volume of both materials:

V_{Pb}=5/11,3=440cm^3

V_{Cu}=5/7,9=630cm^3

And the buoyant forces on elements are:

P_{Pb}=440*1*9,81/1000=4,32N

P_{Pb}=630*1*9,81/1000=6,18N

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a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha
defon

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

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m_1 = 3 kg is the mass of the putty

u_1 = 10 m/s is the initial velocity of the putty (we take its direction as positive direction)

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Re-arranging the equation and substituting the values, we find the  final combined velocity:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s

And the positive sign indicates their final direction is the same as the initial direction of the putty.

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3 years ago
the maximum displacement of a particle, within a wave, above or below its equilibrium position, is called__________
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What would be the answer for this and how?
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Answer:

B. 6 cm

Explanation:

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k = \frac{F}{\Delta x}\\

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k = spring constant of single spring = ?

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Δx = extension = 4 cm = 0.04 m

Therefore,

k = \frac{10\ N}{0.04\ m}\\k =  250\ N/m\\

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

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