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Alika [10]
3 years ago
14

A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A

t the bottom of the incline the cart collides with a solid object and comes to a stop. The distance between the cart and start position is 20 cm.
Velocity Before Velocity After Impulse
0.66 m/s -0.59 m/s 0.287 N*s
1. Find the momentum before and after the collision
2. Next, find the change in momentum
3. Calculate the percent different between the change and momentum and the impulse
Physics
1 answer:
Julli [10]3 years ago
3 0

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

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If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder wil change from 70
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Explanation:

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6 0
3 years ago
17. 53 A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power i
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Answer:

0.337 lb-in

Explanation:

From the law of conservation of angular momentum,

L' = L" where L = initial angular momentum of system and L" = final angular momentum of system

Now L = Iω + Mt where Iω = angular momentum of shaft + wheel and Mt = impulse on system due to couple M.

L' = Iω' + (-Mt) (since the moment about the shaft is negative-anticlockwise)

L' = Iω' - Mt where Iω' = angular momentum of shaft at t' = 0 + wheel and Mt = impulse on system due to couple M in time interval t = 70 s.

L" = Iω"  where Iω" = angular momentum of shaft at t" = 70 s.

Now I = moment of inertia of system = mk² where m = mass of system = W/g where W = weight of system = 6 lb and g = acceleration due to gravity = 32 ft/s². So, m = W/g = 6lb/32 ft/s² = 0.1875 lb-s²/ft and k = radius of gyration = 2 in = 2/12 ft = 1/6 ft.

So, I = mk² = (0.1875 lb-s²/ft) × (1/6 ft)² = ‭0.00521‬ lb-ft-s², ω' = initial angular speed of system = 3600 rpm = 3600 × 2π/60 = 120π  rad/s = 377 rad/s,  ω" = final angular speed of system = 0 rad/s (since it stops), t' = 0 s, f" = 70 s and M = couple on system

So,

Iω' - Mt" = Iω"

Substituting the values of the variables into the equation, we have

Iω' - Mt" = Iω"

0.00521‬ lb-ft-s² ×  377 rad/s - M × 70 s = 0.00521‬‬ lb-ft-s² × 0 rad/s"

0.00521‬ lb-ft-s² ×  377 rad/s - 70M = ‭0

1.964‬ lb-ft-s = 70M

M = 1.964‬ lb-ft-s/70 s

M = 0.0281‬ lb-ft

M = 0.0281 lb × 12 in

M = 0.337 lb-in

6 0
3 years ago
Consider the following isobaric process for air, modelled as a Calorically Perfect Ideal Gas
Readme [11.4K]

These given conditions satisfy the second law of thermodynamics.

As the process is isobaric

So there will be a straight line of P= 200kPa in P-v and P-T planes

P1 = P2 = 100kPa

For perfect ideal gas, v-T plane:

v = (\frac{R}{P}) T

v_{1}  = (\frac{R}{P_{1} }) T_{1} = 287 × 500/200000 = 0.717 m³/kg

v_{2}  = (\frac{R}{P_{2} }) T_{2} = 287 × 600/200000 = 0.861m³/kg

As it is the calorically perfect gas

de = c_{v}dT

Integration on both sides

e2 - e1 = c_{v}(T2 - T1)

           = ( 716.5J/kg/K) (600-500)

           = 71650 J/kg

also,

Tds = de + Pdv

Tds = c_{v}dT +Pdv

For ideal gas

V = RT/P        

dv = Rdt/P - RTdp/P²

Tds = c_{v}dT + Rdt - RTdp/P

ds = (c_{v} + R)dT/T - RdP/P

ds = (c_{v} + c_{p} -c_{v})dT/T - RdP/P

ds = c_{p}dT/T - RdP/P

Integration on both sides

s2 - s1 = c_{p}ln (T2/T1) - R ln (P2/P1)

Since P is constant

s₂ - s₁ = c_{p} ln (T2/T1)

           = 1003.5 ln (600/500)

           = 1003.5 × 0.182

           = 182.95 J/kg/K

w = Pdv

w_{12} = P(v₂ - v₁)

     = 2,00,000 ( 0.861 - 0.717)

     = 28,800 J/kg

de = δq -δw

δq = de + δw

q₁₂ = (e₂ - e₁) +  w₁₂

    =  71,650 + 28,800 = 1,00,450 J/kg

Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.

Let’s check for second law satisfaction.

s₂ - s₁ ≥ q₁₂ / Tₓ

182.95 ≥ 1,00,450 / 600 K

182.95 J/kg/K ≥ 167.41 J/kg/K

Hence this condition satisfies the second law of thermodynamics

Learn more about laws of thermodynamics here brainly.com/question/13164851

#SPJ1

3 0
2 years ago
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