The

Two things that are magnetic but are not magnets

Feliz [49]

Answer:

aluminum and steel are both magnetics

7 0

3 years ago

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Two isolated copper plates, each of area 0.40 m2, carry opposite charges of magnitude 7.08 × 10-10 C. They are placed opposite e

polet [3.4K]

Answer:

The potential difference between the plates is 8 V.

Explanation:

Given that,

Area of plates = 0.40 m²

Charge $q=7.08\times10^{-10}\ C$

Distance = 4.0 cm

We need to calculate the electric field

Using for formula of electric field

$E=\dfrac{2q}{2\epsilon_{0}A}$

Where, q = charge

A = area

Put the value into the formula

$E=\dfrac{7.08\times10^{-10}}{8.85\times10^{-12}\times0.40}$

$E=200\ V/m$

We need to calculate the potential difference between the plates

Using formula of potential difference

$V=E\times d$

Where, E = electric field

d = distance

Put the value into the formula

$V=200\times0.04$

$V=8\ V$

Hence, The potential difference between the plates is 8 V.

3 0

4 years ago

A car goes from a speed of 10m/s to 30m/s over 4s. If the car has a mass of 1000kg, How strong is the force pushing the car forw

dusya [7]

Answer:

i think ww

Explanation:

i think we need to calculate net force which is:

net force=m.a

..nf=1000kg×30m/s

therefore net force will be the answer you get whwn you multiply those two

5 0

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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7 0

4 years ago

The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by

AlekseyPX

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3 0

4 years ago