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3 years ago

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The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algeb

ra, find an equation for an object's speed v after falling a vertical distance h. Do this by equating KE to its change of PE.

1 answer:

Delvig [45]3 years ago

4 0

Answer:

$v = \sqrt{20h}$

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (<em>divide by m throughout</em>)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (<em>substitute g=10</em>)

v²= 20h (<em>×2 on both sides</em>)

v= √20h (<em>square root both sides</em>)

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Goryan [66]

Answer:

<em>The velocity of the truck is 3.33 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If some collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

There are two objects: The m1=4000 Kg car and the m2=6000 Kg truck. The car was moving initially at v1=4 m/s and the truck was at rest v2=0. After the collision, the car moves at v1'=-1 m/s. We need to find the velocity of the truck v2'. Solving for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

Substituting:

$\displaystyle v'_2=\frac{4000*4+6000*0-4000(-1)}{6000}$

$\displaystyle v'_2=\frac{16000+4000}{6000}$

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The velocity of the truck is 3.33 m/s

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Strike441 [17]

Answer : The energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

Explanation :

First we have to calculate the mass defect $(\Delta m)$ .

The balanced reaction is,

$^{232}Th\rightarrow ^{228}Ra+^{4}He$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})$

$\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg$

conversion used : $(1amu=1.66\times 10^{-27}kg)$

Now we have to calculate the energy released.

$Energy=\Delta m\times (c)^2$

$Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=7.974\times 10^{-13}J$

The energy released is $7.974\times 10^{-13}J$

Therefore, the energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

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hoa [83]

Answer:

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A student solving a physics problem for the range of a projectile has obtained the expression

Alika [10]

Answer:

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Explanation:

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So, the range of the projectile is 66.7 meters. Hence, this is the required solution.

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