Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 7, and the turbine inlet temperature is 1200 K. For constant specific heats with k = 1.4 and Cp = 1.005 kJ/kg, calculate the percent thermal efficiency (enter a number only)

Answer:

32.96 MPa

Explanation:

The Poisson ratio of copper is:

μ = 0.355

The Young's modulus of copper is:

E = 117 GPa

The equation for reduction of diameter of a rod is:

D = D0 * (1 - μ*σ/E)

Rearranging:

D = D0 - D0*μ*σ/E

D0*μ*σ/E = D0 - D

D0*μ*σ = E*(D0 - D)

σ = E*(D0 - D) / (D0*μ)

If the diameter is reduced by 0.01 percent

D = 0.9999*D0

σ = E*(D0 - 0.9999*D0) / (D0*μ)

σ = E*(0.0001*D0) / (D0*μ)

σ = 0.0001*E / μ

σ = 0.0001*117*10^9 / 0.355 = 32.96 MPa

This value is below the yield strength, therefore it is valid.

The total kinetic energy of the system is approximately 17128.26 J.T

What is the energy?

Calculate the moment of inertia (I) of the pulley:

= 0.5 * mass * (outer radius² + inner radius²)

= [tex]0.5 * 12.0 kg * ((0.250 m)² + (0.200 m)²)[/tex]

= 1.925 kg * m²

Use the parallel-axis theorem to find I:

I = [tex]1.925 kg * m² + 12.0 kg * (0.231 m)²[/tex]

I = 2.5683 kg * m²

Calculate the kinetic energy of the pulley:

= 0.5 * I * omega²

= [tex]0.5 * 2.5683 kg * m² * (69.0 rad/s)[/tex]

= 6555.63 J

Calculate the linear velocity of cylinder A:

v = outer radius * omega

v =[tex]0.250 m * 69.0 rad/s[/tex]

v = 17.25 m/s

Calculate the kinetic energy of cylinder A:

= 0.5 * mass * v²

= [tex]0.5 * 71.0 kg * (17.25 m/s)²[/tex]

= 10572.63 J

KEtotal = KEpulley + KEcylinder

=  [tex]6555.63 J + 10572.63 J[/tex]

= 17128.26 J

Answer:

(1) [tex]\Delta E = 4845.43 kW[/tex]

(2) [tex]\Delta E_{m} = 5.7319 kW[/tex]

(3) [tex]\Delta E_{t} = 4839.69 kW[/tex]

(4) q = 4839.69 kW[/tex]

Solution:

Using Saturated water-pressure table corresponding to pressure, P = 10 bar:

At saturated temperature, Specific enthalpy of water, [tex]h_{ws} = h_{f} = 762.5 kJ/kg[/tex]

At inlet:

Saturated temperature of water, [tex]T_{sw} = 179.88^{\circ}C[/tex]

Specific volume of water, [tex]V_{wi} = V_{f} = 0.00127 m^{3}/kg[/tex]

Using super heated water table corresponding to a temperature of [tex]600^{\circ}C[/tex] and at  7 bar:

At outlet:

Specific volume of water, [tex]V_{wso} = 0.5738 m^{3}/kg[/tex]

Specific enthalpy of water, [tex]h_{wo} = 3700.2 kJ/kg[/tex]

Now, at inlet, water's specific enthalpy is given by:

[tex]h_{i} = C_{p}(T - T_{sw}) + h_{ws}[/tex]

[tex]h_{i} = 4.187(110^{\circ} - 179.88^{\circ}) + 762.5[/tex]

[tex]h_{i} = -292.587 + 762.5= 469.912 kJ/kg[/tex]

(1) Now, the change in combined thermal energy and work flow is given by:

[tex]\Delta E = E_{o} - E_{i}[/tex]

[tex]\Delta E = m(h_{wo} - h_{i})[/tex]

[tex]\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW[/tex]

(2) The mechanical energy can be calculated as:

velocity at inlet, [tex]v_{i} = \rho A V_{wi}[/tex]

[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]

[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]

[tex]v_{i} = \frac{1.5\times 0.00127}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]

[tex]v_{i} = 0.542 m/s[/tex]

Similarly,, the velocity at the outlet,

[tex]v_{o} =  \frac{1.5\times 0.57378}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]

[tex]v_{o} =  276.099 m/s[/tex]

Now, change in mechanical energy:

[tex]\Delta E_{m} = E_{mo} - E_{mi}[/tex]

[tex]\Delta E_{m} = m[(\frac{v_{o}^{2}}{2} + gz_{o}) - (\frac{v_{i}^{2}}{2} + gz_{i})][/tex]

[tex]\Delta E_{m} = 1.5[(\frac{276.099^{2}}{2} + 9.8(z_{o} - z_{i}) - (\frac{0.542^{2}}{2}][/tex]

[tex]\Delta E_{m} = 57319 J = 5.7319 kW[/tex]

(3) The total energy of water is given by:

[tex]\Delta E_{t} = E - E_{m} = 4845.43 - 5.7319 = 4839.69 kW[/tex]

(4) The rate of heat transfer:

q = [tex]\Delta E_{t} = 4839.69 kW[/tex]

Answer:

Acceleration in circular path will be 9

Explanation:

We have given speed of the particle in circular path = 6 m/sec

Radius of the circular path = 4 m

We have to find the centripetal acceleration [tex]a_c[/tex]

We know that centripetal acceleration is given by [tex]a_c=\frac{v^2}{r}=\frac{6^2}{4}=9[/tex]

As in question it is given that don't include the unit

So acceleration will be 9

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So [tex]250lb-ft=250\times 4.45N\times 0.348M=338.95N-m[/tex]

So 250 lb-ft = 338.95 N-m

Answer:

The correct answer is 'velocity'of liquid flowing out of an orifice is proportional to the square root of the 'height'  of liquid above the center of the orifice.

Explanation:

Torricelli's theorem states that

[tex]v_{exit}=\sqrt{2gh}[/tex]

where

[tex]v_{exit}[/tex] is the velocity with which the fluid leaves orifice

[tex]h[/tex] is the head under which the flow occurs.

Thus we can compare the given options to arrive at the correct answer

Velocity is proportional to square root of head under which the flow occurs.

Answer:

Q(h=200)=0.35W

Q(h=3000)=5.25W

Explanation:

first part h=200W/Km^2

we must use the convection heat transfer equation for the chip

Q=hA(Ts-T∞)

h= convective coefficient=200W/m2 K

A=Base*Leght=5mmx5mm=25mm^2

Ts=temperature of the chip=85C

T∞=temperature of coolant=15C

Q=200x2.5x10^-5(85-15)=0.35W

Second part h=3000W/Km^2

Q=3000x2.5x10^-5(85-15)=5.25W

Explanation:

In shaping work piece will be stationary and tool will reciprocates,but on the other hand in planning work piece will reciprocates and tool will be stationary.Shaping is used for small work piece and planning is used for large work piece.Both shaping and planning are not continuous cutting process,cutting action take place in forward stroke and return stroke is idle stroke.The velocity of return stroke is much more than the forward stroke.

To determine the instantaneous velocity and acceleration of a particle described by the position function s(t), one needs to calculate the first and second derivatives of the position function and evaluate them at the given time, t = 8 s. The average velocity and speed require the change in position over a specific time interval, which is not provided in the question.

The position of a particle along a straight-line path is given by the equation s = (t3 - 6t2 - 15t + 7) feet, where t is time in seconds. To find the particle's instantaneous velocity and instantaneous acceleration at t = 8 s, we need to take the first and second derivatives of the position function with respect to time, respectively.

The first derivative of s with respect to t gives us the velocity v(t), and the second derivative gives us the acceleration a(t). At t = 8 s, the velocities and accelerations can be calculated by plugging in the value of t into those derivatives.

To determine the average velocity and average speed, we take the change in position over the change in time interval for the specific time range provided.

Note: The question lacks sufficient specific information to calculate these values as the time interval for the average velocity and average speed is not provided. However, the general process of calculation has been explained.

Acceleration, instantaneous velocity, and particle position at different times are essential concepts in physics and particle motion analysis.

Acceleration is the rate of change of velocity with respect to time. In the given scenarios, acceleration is provided in terms of a function of time or as a constant value.

Instantaneous velocity is the velocity of the particle at a specific moment. It can be calculated by taking the derivative of the position function with respect to time.

Position of the particle at different times can be found by substituting the respective time values into the position function.

Answer:

True

Explanation:

Engineering controls are those techniques used to reduce or eliminate hazards of any condition, thereby protecting the workers.

These are mostly products that act as barriers between the worker and the hazard. This may include machinery or equipment. The common engineering controls used are glovebox, biosafety cabinet, fume hood, vented balance safety enclosure, HVAC system, lockout-tagout, sticky mat and rupture disc.

Answer:

Lets take [tex]Q_2[/tex] heat transfer take place from 500 K reservoir to device and  [tex]Q_3[/tex] from device to  300 K reservoir.

From the energy conservation we can say that

[tex]400+Q_2=100+Q_3[/tex]  

[tex]300=Q_3-Q_2[/tex]   -----1

For reversible process

[tex]\dfrac{400}{1000}+\dfrac{Q_2}{500}-\dfrac{Q_3}{300}=0[/tex]  

[tex]5Q_3-3Q_2=600[/tex]        ----2

By solving above two equation

[tex]Q_3=-150 KJ,Q_2=-450KJ[/tex]

But here sign come negative it means that

[tex]Q_2[/tex] heat transfer take place from device to 500 K reservoir and [tex]Q_3[/tex] from 300 K reservoir to device.

In this exercise we have to use the knowledge of heat transfer to calculate the heat transferred to the other two reservoirs will be:

The magnitude of the Q2 is -450 in the out direction while the Q3 is -150 in the inward direction.

From the information given in the statement, we have that:

knowing that from conservation we can say:

[tex]400+Q_2=100+Q_3\\300=Q_3-Q_2\\\frac{400}{1000}+\frac{Q_2}{500}-\frac{Q_3}{300}=0\\[/tex]

So solving we have:

[tex]Q_3=-150KJ\\Q_2=-450KJ[/tex]

See more about heat transfer at brainly.com/question/12107378

Answer:

a) on mars W=1454.7N

b)on earth W=3825.9N

Explanation:

The weight of any body with mass is given by the following equation

W=mg

where

m=mass

g=gravity

W=weight

Remember that the weight is expresed in Newton and the units are kgm/s^2

A)weight on the surface of mars

W=(390kg)(3.73m/s^2)=1454.7N

b) on earth

W=(390kg)(9.81m/s^2)=3825.9N

Answer:

You can't divide by zero

Explanation:

The error appears when you divide each side by (a - b). If a = b, then (a - b) = 0 and you can't divide each side by 0. Moreover, the equation before division, that is, (a + b)(a − b) = b(a − b) is true after replacing (a - b) = 0  because it gives 0 = 0.

The error in the proof lies in the step where we divided both sides by (a - b) . Since [tex]\( a = b \), \( (a - b) \)[/tex] becomes 0, making the division by 0 undefined.

The error occurs when dividing both sides by (a - b) . Since ( a = b ), ( (a - b)  becomes 0. Division by 0 is undefined, leading to the invalid conclusion.

To elaborate, dividing both sides by (a - b)  in the equation (a + b)(a - b) = b(a - b) results in:

[tex]\[ \frac{(a + b)(a - b)}{(a - b)} = \frac{b(a - b)}{(a - b)} \]\[ \frac{(a + b)\cancel{(a - b)}}{\cancel{(a - b)}} = \frac{b\cancel{(a - b)}}{\cancel{(a - b)}} \]\[ a + b = b \][/tex]

This step assumes  (a - b)  is not equal to zero, leading to the false conclusion that ( a + b = b ), which is only true when  a  and ( b ) are equal.

Therefore, the proof incorrectly concludes that ( 2 = 1 ) due to the division by zero, highlighting the importance of avoiding such mathematical errors.

Answer with Explanation:

i) Boundary layer thickness: It is the thickness of the boundary layer formed around an object that is placed in the path of a flowing viscous fluid.The boundary layer thickness is the thickness up to which the effect of the object on the flow can be felt. When a viscous flowing fluid encounters an object in it's path of flow, the flowing fluid forms a thin layer of fluid over the object and this layer of fluid is known as boundary layer. This is a phenomenon only observed in the viscous fluids. As shown in the below figure a uniform flow of a viscous fluid encounters a plate, as we can see the thickness of the boundary layer goes on increasing as we move away from the leading edge of the plate the thickness of the boundary layer at any position is termed as boundary layer thickness.

ii) Boundary layer transition: It is the transition of the flow from a laminar nature to fully developed turbulent flow as it moves over an object. It occurs due to change in the Reynolds number of the flow as the effect of boundary layer increases as we move away from the leading edge of the object.  

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = [tex]\frac{5}{18}\times50[/tex] = 13.89 m/s

Now,

We have the relation

[tex]\tan\theta=\frac{v^2}{gR}[/tex]

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

[tex]\tan\theta=\frac{13.89^2}{9.81\times100}[/tex]

or

[tex]\tan\theta=0.1966[/tex]

or

θ = 11.125°

Answer:

The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.

Explanation:

The net force over the window is calculated by multiplying the difference in pressure by the area of the window:

F = Δp*A

The pressure inside the plane is around 1 atm, hence the difference in pressure is:

Δp = 1atm - 0.35 atm = 0.65 atm

Expressing in the EE unit system:

Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2

Replacing in the force:

F = 9.55 lbf/in^2 * 80 in^2  = 764 lbf

For the international unit system, we re-calculate the window's area and the difference in pressure:

A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2

Δp =  0.65 atm * 101325 Pa  = 65861 Pa  = 65861 N/m^2

Replacing in the force:

F = 65861 N/m^2  *0.0516 m^2  = 3398 N

Answer:

The mass in:

1) Grams =[tex]377765.5\times 10^{3}grams[/tex]

2) Kilograms =[tex]377765.5kilograms[/tex]

3)Tonnes =[tex]377.7655tonnes[/tex]

4) Megatonnes =[tex]0.378megatonnes[/tex]

Explanation: The given mass of the aircraft in pounds is 833,000 pounds.

Part 1)

Since we know that 1 pound equals 453.5 grams thus by ratio we have

833,000 pounds =[tex]833000lb\times 453.5\frac{g}{lb}=377765.5\times 10^{3}grams[/tex]

Part 2)

Since we know that 1000 grams equals 1 kilogram

thus the above mass in kilograms equals[tex]\frac{377765.5\times 10^{3}}{1000}=377765.5kg[/tex]

Part 3)

Since there are 1000 kilograms in 1 tonne

thus the given mass is converted into tonnes as

[tex]mass_{tonnes}=\frac{377765.5kg}{1000}=377.765tonnes[/tex]

Part 4)

Since 1 Mega tonne(Mton) equals 1000 tonnes thus the given mass is converted into mega tonnes as

[tex]mass_{M\cdot tonn}=\frac{377.7655tonnes}{1000}=0.378Megatonnes[/tex]

Answer:

The temperature in degree Celsius will be -238.7055°C

Explanation:

We have given the substance temperature = 62°R

We have to convert degree Rankine to degree Celsius

For conversion from Rankine to Celsius we use formula

[tex]T_C=(T_R-491.67)\times\frac{5}{9}[/tex]

So [tex]T_C=(62-491.67)\times\frac{5}{9}[/tex]

[tex]T_C=-238.7055^{\circ}C[/tex]

So temperature in degree Celsius will be -238.7055°C  

After calculation i got -238.7055°C but in option this is not given

The temperature of the substance will be "94.4°C". To understand the calculation, check below.

According to the question,

Substance temperature, T°R = 62

or,

T°C = (T°R - 491.67) × [tex]\frac{5}{9}[/tex]

By substituting the values,

      = -238.706

If we take the value,

T°C = (662 - 491.67) × [tex]\frac{5}{9}[/tex]

      = 94.62°C or,

      = 94.4°C

Thus the above response "Option D" is correct.

Find out more information about temperature here:

https://brainly.com/question/16559442

Answer:

The heat from the sun melted it

Explanation:

If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

Answer:

Modulus of resilience will be [tex]3216942.308j/m^3[/tex]

Explanation:

We have given yield strength [tex]\sigma _y=818MPa[/tex]

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience [tex]=\frac{\sigma _y^2}{2E}[/tex], here [tex]\sigma _y[/tex] is yield strength and E is elastic modulus

Modulus of resilience [tex]=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3[/tex]  

Answer:

The correct answer is option 'a': Black

Explanation:

As we know that for an object which is black in color it absorbs all the electromagnetic radiation's that are incident on it. Thus if we need to transfer energy to an object by radiation the most suitable color for the process  is black.

In contrast to black color white color is an excellent reflector, reflecting all the incident radiation that may be incident on it hence is the least suitable material for radiative heat transfer.

Explanation:

Step1

Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.

Step3

Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.

The expression for factor of safety for ductile material is given as follows:

[tex]FOS=\frac{\sigma_{yp}}{\sigma_{a}}[/tex]

Here,[tex]\sigma_{f}[/tex] is yield stress and [tex]\sigma_{a}[/tex] is allowable stress.

The expression for factor of safety for brittle material is given as follows:

[tex]FOS=\frac{\sigma_{ut}}{\sigma_{a}}[/tex]

Here,[tex]\sigma_{ut}[/tex] is ultimate stress and [tex]\sigma_{a}[/tex] is allowable stress.

Answer:

option D is correct

Explanation:

1) for a cylinder with radius is[tex]\frac{80}{\pi}[/tex]

lateral surface area of cylinder is [tex]2\pi rh[/tex]

                                                   [tex]= 2\pi \frac{80}{\pi}*120[/tex]

                                                   = 160* 120

2) fro square prism with side 40 mm

  lateral surface area = 4ah

                                   = 4*40 * 120

                                   = 160*120

3)for hexagonal with side 20 mm

lateral surface area = 6ah

                                = 6*20*160

                                 =120* 160

therefore option D is correcrt

Answer:

6.8 kJ

Explanation:

p1 = 95 kPa

T1 = 300 K

T3 = 2100 K

m = 12 g

Ideal gas equation:

p * v = R * T

v = R * T / p

R for air is 0.287 kJ/(kg K)

v1 = 0.287 * 300 / 95 = 0.9 m^3/kg

v2 = v1 / cr

v2 = 0.9 / 18 = 0.05 m^3/kg

Assuming an adiabatic compression

p*v^k = constant

k is 1.4 for air

p1 * v1 ^ k = p2 * v2 ^ k

p2 = p1 * (v1 / v2) ^k

p2 = p1 * cr^k

p2 = 95 * 18^1.4 = 5.43 MPa

p1*v1/T1 = p2*v2/T2

T2 = p2*v2*T1/(p1*v1)

T2 = 5430 * 0.05 * 300 / (95 * 0.9) = 952 K

The first principle of thermodynamics

Q = W + ΔU

Since this is an adiabatic process Q = 0

W = -ΔU

W1-2 = -m * Cv * (T2 - T1)

The Cv of air is 0.72 kJ/kg

W1-2 = -0.012 * 0.72 * (952 - 300) = -5.63 kJ

Next the combustion happens and temperature increases suddenly.

v3 = v2 = 0.05 m^3/kg

T2 * p2^((1-k)/k) = T3 * p3^((1-k)/k)

p3 = p2 * (T2/T3)^(k/(1-k)

p3 = 5430 * (952/2100)^(1.4/(1-1.4) = 86.5 MPa

The work is zero because the piston doesn't move.

Next it expands adiabatically:

v4 = v1 = 0.9 m^3/kg

T * v^(k-1) = constant (adiabatic process)

T3 * v3^(k-1) = T4 * v4^(k-1)

T4 = T3 * (v3 / v4)^(k-1)

T4 = 2100 * (0.05 / 0.9)^(1.4-1) = 661 K

p3*v3/T3 = p4*v4/T4

p4 = p3*v3*T4/(v4*T3)

p4 = 86500*0.05*661/(0.9*2100) = 1512 kPa

L3-4 = -m * Cv * (T4 - T3)

L3-4 = -0.012 * 0.72 * (661 - 2100) = 12.43 kJ

Net work:

L1-2 + L3-4 = -5.63 + 12.43 = 6.8 kJ

Answer:

Hooke's law is a principle of physics that states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. That is: F = kX, where k is a constant factor characteristic of the spring: its stiffness, and X is small compared to the total possible deformation of the spring.

Explanation:

Answer:

total weight of liquid = 4788.25 N or 488.09 kg

Explanation:

given data

total volume = 0.5 m³

volume of ethanol = 10 %  of volume = 0.10 × 0.5 = 0.05 m³

volume of water = 90 % at 25 °C of volume = 0.90 × 0.5 = 0.45 m³

to find out

weight of the liquid

solution

we know that density of water at 25  is 997 kg/m³

and density of ethanol is 789 kg/m³

so weight of water is = density × volume × g

put here value and we take g = 9.81

weight of water is = 997 × 0.45 × 9.81

weight of water = 4401.25 N     ......................1

weight of ethanol is = density × volume × g

put here value and we take g = 9.81

weight of ethanol is = 789 × 0.05 × 9.81

weight of ethanol = 387.00 N       ...............2

so total weight of liquid = sum of equation 1 add 2

total weight of liquid =  4401.25 + 387

total weight of liquid = 4788.25 N or 488.09 kg

Answer:

The correct answer is option 'b': 9.9 m/s

Explanation:

We know that for an ideal fluid the velocity of exit from a tank with the height of water 'h' is given by Torricelli's Law as

[tex]v=\sqrt{2gh}[/tex]

where,

'g' is acceleration due to gravity

'h' is the level of water

Applying the given values we obtain velocity as

[tex]v=\sqrt{2\times 9.81\times 5}=9.90m/s[/tex]

Answer:

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t  - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

Answer:

Final Velocity (Vf)= 139.864 ft/s

Time (t)= 4,34 s

Explanation:

This is a free fall problem, to solve it we will apply free  fall concepts:

In a free fall the acceletarion is gravity (g) = 9,81 m/s2, if we convert it to ft/s^2 = g= 32.174 ft/s^2

where h is height (304 ft in this case).

Vo =0 since the hammer wasn't moving when it stared to fall

Then Vf^2= 0 + 2* 32.174 ft/s^2 *304 ft

          Vf^2= 19,561.8224  ft^2/s^2

          Vf=[sqrt{19561.8224 ft^2/s^2}

          Vf=139.864 ft/s

Time t= (Vf-Vo)/g => (139.864 ft/s-0)/32.174 ft/s^2 = 4.34 sec

Good luck!