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3 years ago

12

What are two results of animal regeneration?

2 answers:

attashe74 [19]3 years ago

6 0

That will most likely become extinct

laiz [17]3 years ago

3 0

MOre animal population and less eligible of becoming extinct!

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An object's starting position is at point A and its final position is at point B.

salantis [7]

B- we would need the time to fulfill the formula distance = speed x time

4 0

3 years ago

The density of aluminum is 2.7 × 103 kg/m3 . the speed of longitudinal waves in an aluminum rod is measured to be 5.1 × 103 m/s.

andrey2020 [161]

<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2. So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density. So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave. Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>

5 0

3 years ago

What is the density of a piece of wood with a mass of 25 kg<br> and a volume of 0.0385 m³?

Orlov [11]

Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

$p = \frac{mass \: (in \: kg)}{volume \: (in \: {m}^{3}) }$

We can substitute the givenmass and volume to find density of the object.

$p = \frac{25kg}{0.0385 {m}^{3} } \\ = 649kg \: per \: {m}^{3}$

Therefore the density of this object is 649kg/m^3.

7 0

2 years ago

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi

antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

$Kc=510=1/2*Ic*\omega c^2$

$\omega c=\sqrt{510*2/Ic}$

Where the inertia is given by:

$Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2$

Replacing this value:

$\omega c=106.46rad/s$

Speed of the block will therefore be:

$V_b=\omega_c*R_c=106.46*0.15=15.969m/s$

By conservation of energy:

Eo = Ef

Eo = 0

$Ef = 510+1/2*m_b*V_b^2-m_b*g*h$

So,

$0 = 510+1/2*m_b*V_b^2-m_b*g*h$

Solving for h we get:

h=16.67m

3 0

3 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em /> $m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em>so since the bottle is at rest firstly, therefore </em> $v_{i2} =0$ <em />

<em /> $m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em /> $m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ <em> </em><em>equation 1</em>

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

<em /> $m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$ <em />

<em>collect the like terms</em>

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

<em />

6 0

3 years ago

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