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Debora [2.8K]
3 years ago
14

Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other.

They have a head-on collision and bounce off away from each other. Assuming this is a perfectly elastic collision, what will be the approximate kinetic energy of the system after the collision?
A. 1.6 × 105 joules
B. 2.5 × 105 joules
C. 1.2 × 103 joules
D. 2.5 × 103 joules ...?
Physics
1 answer:
juin [17]3 years ago
5 0
A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:

KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A
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Scalars are used to solve projectile motion problems because they allow the analysis of one direction at a time for two-dimensio
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3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a
inessss [21]

Answer:

The average angular acceleration is  \alpha =125.487 rad /s^2

Explanation:

From the question we are told that

  From the question we are told that

        The length of the bat is l = 0.85m  \

         The initial linear velocity is  u = 0 m/s

         The time is  t = 0.15s

         The velocity at t is  v = 16 m/s

  Generally average  angular acceleration is mathematically represented as

                \alpha  = \frac{w_f - w_o}{t}

        Where w_f is the finial angular velocity which is mathematically evaluated as  

            w_f = \frac{v}{l}

                  w_f = \frac{16}{0.85}

                        = 18.823 rad/s

 and w_o is the initial angular velocity which is zero since initial linear velocity is zero

               So

                         \alpha  = \frac{18.823 - 0}{0.15}

                               \alpha =125.487 rad /s^2

5 0
3 years ago
Describe what a hydrogen bond is. 6th grade answer
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Answer:

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3 years ago
Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
3 years ago
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