Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

Answers

Answer 1

Answer:

Answer:

The exponent A in the equation is 3.

Explanation:

v = a^2 t^ A /x

$v = (a^2t^A)/(x) \n\nvx = a^2t^A\n\n((L)/(T))(L) = ((L)/(T^2))^2(T)^A\n\n (L^2)/(T)= ((L^2)/(T^4))(T)^A\n\n (L^2)/(T) *(T^4)/(L^2) = (T)^A\n\nT^3 = T^A\n\n(T^3)/(T^3) = (T^A)/(T^3)\n\nT^(3-3) = T^(A-3)\n\n3-3 = A-3\n\n0 = A-3\n\nA = 3$

Therefore, the exponent A in the equation is 3.

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A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until the volume quadruples and the cylinder contains saturated vapor only. Determine (a) the volume of the cylinder, (b) the final temperature.

Answers

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^(\circ)$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4* 0.001157$

$V_1=1.6198* 10^(-3)\ m^3$

Final Volume $V_2=4V_1$

$V_2=4* (1.6198* 10^(-3))$

$V_2=6.4792* 10^(-3)\ m^3$

Specific volume at this stage

$\nu _2=(V_2)/(m)$

$\nu _2=(6.4792* 10^(-3))/(1.4)$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^(*)+(T_2^(*)-T_1^(*))/(\alpha _2^(*)-\alpha _1^(*))* (\alpha _2-\alpha _1^(*))$

$T_2=370^(\circ)+(373.95-370)/(0.003106-0.004953)* (0.004628-0.004953)$

$T_2=370.7^(\circ) C$

Final answer:

The problem in the question is solved using the principles of thermodynamics. The volume of the device after the heat transfer is 6311.2 cm³. The final temperature inside the cylinder, when the water reached the state of saturated vapor, is approximately 240°C.

Explanation:

The subject question is a thermodynamics problem; more specifically dealing with changes of state, volume, and temperature in a system under certain conditions.

For solving part (a), one would first need to find the specific volume (v) at the initial state, which is saturated liquid at 200°C. Looking up in the property tables, we see that v = 1.127 cm³/g for saturated water at 200°C. Then, the initial volume (V) is mass times specific volume, so V = 1.4 kg x 1.127cm³/g x 1000g/kg = 1577.8 cm³. Because volume quadrupled, the final volume is 4 x 1577.8 cm³ = 6311.2 cm³.

For part (b), at the final state, the water is a saturated vapor. The specific volume at the final state is the final volume divided by the mass, which equals to 6311.2 cm³ / 1.4kg / 1000g/kg = 4.507 cm³/g. Look this value up in the property table to find the corresponding temperature. We get a final temperature of about 240°C.

Learn more about Thermodynamics here:

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2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.

Answers

Answer:

The final temperature is 50.8degrees celcius

Explanation:

Pls refer to attached handwritten document

Answer: 50.63° C

Explanation:

Given

Length of heater, L = 200 mm = 0.2 m

Diameter of heater, D = 15 mm = 0.015 m

Thermal conductivity, k = 5 W/m.K

Power of the heater, q = 25 W

Temperature of the block, = 35° C

T1 = T2 + (q/kS)

S can be gotten from the relationship

S = 2πL/In(4L/D)

On substituting we have

S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)

S = 1.2568 / In 53.33

S = 1.2568 / 3.98

S = 0.32 m

Proceeding to substitute into the main equation, we have

T1 = T2 + (q/kS)

T1 = 35 + (25 / 5 * 0.32)

T1 = 35 + (25 / 1.6)

T1 = 35 + 15.625

T1 = 50.63° C

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?

Answers

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

$F=(mv^2)/(r)$

$v=\sqrt{(Fr)/(m)}$

$v=\sqrt{(0.025* 0.29)/(0.01)}$

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

A basketball has a mass of 575 g. Moving to the right and heading downward at an angle of 31° to the vertical, it hits the floor with a speed of 4 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 31° to the vertical. What was the momentum change ? (Take the axis to be to the right and the axis to be up. Express your answer in vector form.)

Answers

Answer:

Taking the x axis to the right and the y axis to be up, the total change of momentum is $\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}$

Explanation:

The momentum $\vec{p}$ is given by:

$\vec{p} = m \ \vec{v}$

where m is the mass and $\vec{v}$ is the velocity. Now, taking the suffix i for the initial condition, and the suffix f for the final condition, the change in momentum will be:

$\Delta \vec{ p} = \vec{p}_f - \vec{p}_i$

$\Delta \vec{ p} = m \ \vec{v}_f - m \ \vec{v}_i$

$\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)$

As we know the mass of the ball, we just need to find the initial and final velocity.

Knowing the magnitude and direction of a vector, we can obtain the Cartesian components with the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector and θ is the angle measured from the x axis.

Taking the x axis to the right and the y axis to be up, the initial velocity will be:

$\vec{v}_i = 4 (m)/(s) ( \ cos ( - (90 \ °- 31 \°)) , sin( - (90 \ ° - 31\°) ) ) =$

where minus sign appears cause the ball is going downward, and we subtracted the 31 ° as it was measured from the y axis

So, the initial velocity is

$\vec{v}_i = 4 (m)/(s) ( \ cos ( - 59 \°) , sin( - 59 \°)) =$

$\vec{v}_i = ( \ 2.0601 \ (m)/(s) , - 3.4286 (m)/(s)) =$

The final velocity is

$\vec{v}_i = 4 (m)/(s) ( \ cos ( 90 \ °- 31 \°) , sin( 90 \ ° - 31\°)) =$

$\vec{v}_i = 4 (m)/(s) ( \ cos ( 59 \°) , sin( 59 \°)) =$

$\vec{v}_i = ( \ 2.0601 \ (m)/(s) , 3.4286 (m)/(s)) =$

So, the change in momentum will be

$\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)$

$\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ (m)/(s) , 3.4286 (m)/(s) - ( \ 2.0601 \ (m)/(s) , - 3.4286 (m)/(s)))$

$\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ (m)/(s) - \ 2.0601 \ (m)/(s), 3.4286 (m)/(s) + 3.4286 (m)/(s))$

$\Delta \vec{ p} = 0.575 \ kg (\ 0 , 2 * 3.4286 (m)/(s) )$

$\Delta \vec{ p} = 0.575 \ kg * 2 * 3.4286 (m)/(s) \hat{j}$

$\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}$

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.1 m/s2 while Kathy maintains an acceleration of 4.99 m/s. 2 (a) Find the time at which Kathy overtakes Stan. s from the time Kathy started driving (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him. Kathy m/s Stan m/s

Answers

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement

t₁ : Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁: Stan acceleration

d₂: car displacement

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂: Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ = 1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1=equation 2

1.55*t₁² = 2.495* t₂² , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² =2.495* t₂²

1.55* (t₂² +2t₂+1) =2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0 Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁ t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂

vf₂ =0+4.99* 3.87

vf₂ = 19.31m/s : Kathy's final speed

An F-35 stealth jet takes off from the aircraft carrier Ronald Reagan. Starting from rest, the jet accelerated with a constant acceleration of 55.3 m/s2 along a straight line on the deck. What is the displacement of the jet when it reaches a speed of 181 m/s?

Answers

Answer:

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

Explanation:

Hi there!

The equation of position and velocity of an object traveling with constant acceleration along a straight line are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the object at time t.

If we place the origin of the frame of reference at the point where the jet starts moving, then, x0 = 0. Since the jet starts from rest, v0 is also zero. Then the equations get reduced to the following:

x = 1/2 · a · t²

v = a · t

We know the acceleration and the final velocity of the jet. So, using the equation of velocity, we can find the time it takes the jet to reach that velocity. Then, we can calculate the position of the jet at that time. Since the initial position is zero, the final position of the jet will be equal to the displacement (because displacement = final position - initial position).

v = a · t

v/a = t

181 m/s / 55.3 m/s² = t

t = 3.27 s

The final position of the jet will be:

x = 1/2 · a · t²

x = 1/2 · 55.3 m/s² · (3.27 s)²

x = 296 m

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

To find displacement using constant acceleration,

we can use the following equation: