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3 years ago

After a 65 newton weight has fallen freely from rest a vertical distance of 5.3 meters, the kinetic energy of the weight is

Engineering

1 answer:

inysia [295]3 years ago

3 0

Answer:

The kinetic energy of the weight is 344.5 J

Explanation:

Given that:

Force = F = 65 newton

distance = d = 5.3 meters

We have to find change in kinetic energy ΔK.E

Now we know that, initially kinetic energy was 0 So the formula we use will be:

Work done = Change in kinetic energy

Mathematically,

W = ΔK.E

As we know W = F . d and ΔK.E = K.E(final) - K.E(initial)

So by putting values:

F . d = K.E(final) - K.E(initial)

F . d = K.E(final)

As K.E(initial) is 0 so by putting values of F and d

(65)* (5.3) = K.E(final)

344.5 J = K.E(final)

So the change in K.E will also be 344.5 J

i hope it will help you!

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A carbon resistor has a resistance of 976 ohms at 0 degrees C. Determine its resistance at 89 degrees C

nignag [31]

Answer:

1028.1184 Ohms

Explanation:

Given the following data;

Initial resistance, Ro = 976 Ohms
Initial temperature, T1 = 0°C
Final temperature, T2 = 89°C

Assuming the temperature coefficient of resistance for carbon at 0°C is equal to 0.0006 per degree Celsius.

To find determine its new resistance, we would use the mathematical expression for linear resistivity;

$R_{89} = R_{0} + R_{0}(\alpha T)$

Substituting into the equation, we have;

$R_{89} = 976 + 976*(0.0006*89)$

$R_{89} = 976 + 976*(0.0534)$

$R_{89} = 976 + 52.1184$

$R_{89} = 1028.1184 \ Ohms$

5 0

3 years ago

Sketches are a very efficient way to share ideas. True False

timurjin [86]

Answer:

yes

Explanation:

4 0

3 years ago

Read 2 more answers

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat to the surroundings.

8 0

3 years ago

A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

$m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s$

$h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg$

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

$m_1+m_2+m_3=m$

m=1+1.5+2 Kg/s

m=4.5 Kg/s

Now from energy conservation

$m_1h_1+m_2h_2+m_3h_3=mh$

By putting the values

$1\times 100+1.5\times 120+2\times 500=4.5\times h$

So h=284.44 KJ

4 0

3 years ago

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm

aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

$T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=14.5MPa

$T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=10.3MPa

7 0

3 years ago