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3 years ago

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A bullet with mass m = 5.21 g is moving horizontally with a speed v = 443 m/s when it strikes a block of hardened steel with mas

s M = 14.8 kg (initially at rest). The bullet bounces off the block in a perfectly elastic collision. (a) What is the speed (m/s) of the block immediately after the collision?

1 answer:

AlladinOne [14]3 years ago

6 0

Answer:

0.312 m/s

Explanation:

Elastic collisions conserve momentum and kinetic energy

The velocity of the center of mass will not change. It continues at

0.00521(443) / 14.80521 = 0.155893... ≈ 0.156 m/s

To conserve kinetic energy we can think of the center of mass (CoM) as an ideal spring returning to each mass that strikes it an identical speed of collision in the opposite direction.

The CoM sees the target approach at - 0.156 and will see it depart at 0.156 m/s

A ground based observer sees the target depart at the velocity of the CoM plus the relative velocity .

v = 0.156 + 0.156 = 0.312 m/s

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg

bekas [8.4K]

Answer:

$5.3\times 10^3 N/C$

Explanation:

We are given that

Distance between plates=d=2.2 cm= $2.2\times 10^{-2} m$

$1 cm=10^{-2} m$

$\sigma=47nC/m^2=47\times 10^{-9}C/m^2$

Using $1 nC=10^{-9} C$

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

$E=\frac{\sigma}{2\epsilon_0}$

$E_1=\frac{\sigma}{2\epsilon_0}$

$E_2=\frac{\sigma}{2\epsilon_0}$

$E=E_1+E_2$

$E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Substitute the values

$E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}$

$E=5.3\times 10^3 N/C$

Hence, the magnitude of E in the region between the plates= $5.3\times 10^3 N/C$

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3 years ago

True or False along a divergent boundary, two plates slip past each other,moving in opposite directions.

kiruha [24]

False....................................

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3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

A)

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

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SOVA2 [1]

A nuclear power plant is defined as a thermal power station where electricity is generated by the help of sustainable nuclear fission.

A nuclear power plant needs nuclear reactor as the thermal energy generator .The reactor has various parts where fissionable radioactive material will undergo nuclear fission .Through this process high amount of thermal energy is released which is allowed to water to generate steam.

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