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noname [10]
3 years ago
10

A car is moving at 20 m/s. The driver steps on the gas pedal and accelerates the car for 4 seconds at an acceleration of 4 m/s2.

How fast is the car moving after it accelerates?
Physics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

The final velocity of the car is 36 m/s.

Explanation:

Given;

initial velocity of the car, u = 20 m/s

time of the car acceleration, t = 4 s

acceleration of the car, a = 4 m/s²

the final velocity of the car is calculated as;

v = u + at

where;

v is the final velocity of the car

v = 20 + (4 x 4)

v = 36 m/s

Therefore, the final velocity of the car is 36 m/s.

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Queremos diseñar un montacargas que pueda subir con una rapidez de 12 km/h una mas 700 kg hasta 40 m de altura en un minuto. Cal
Mariulka [41]

Answer:

a) El trabajo realizado es de 274,680 J

b) La potencia de la carretilla elevadora es de 4578 Watts.

c) La energía cinética del montacargas es de 3.888.\overline 8 J

d) La energía potencial del montacargas es de 274.680 Joules.

e) La energía mecánica de la carretilla elevadora 278,568.\overline 8 J

Explanation:

a) Los parámetros dados son;

La velocidad de la carretilla elevadora, v = 12 km / h = 10/3 m / s

La masa que debe levantar la carretilla elevadora, m = 700 kg

La altura a la que se levantará la masa, h = 40 m

El trabajo realizado, W = Fuerza, F × Distancia, h

 La fuerza, F aplicada = El peso de la carga = Masa, m × Gravedad, g

Donde 'g' es la aceleración debida a la gravedad ≈ 9,81 m / s²

∴ Trabajo realizado, W = 700 kg × 9,81 m / s² × 40 m = 274,680 J

b) El tiempo que se tarda en subir 40 m = 1 minuto = 60 segundos

∴ Potencia = Trabajo / tiempo

Por lo tanto, la potencia del montacargas, P = 274,680 J / (60 s) = 4578 Watts

c) Energía cinética, K.E. = 1/2 · m · v²

La energía cinética de la carretilla elevadora, K.E. se da como sigue;

Carretilla elevadora K.E. = 1/2 × 700 kg × (10/3 m / s) ² = 3.888.\overline 8 J

d) La energía potencial del montacargas a 40 m, P.E. = m · g · h

∴ P.E. = 700 kg × 9,81 m / s² × 40 m = 274,680 Julios

e) La energía mecánica, M.E. = P.E. + K.E.

∴ M.E. = 3.888.\overline 8 J + 274,680 J = 278,568.\overline 8 J

La energía mecánica de la carretilla elevadora, M.E.= 278,568.\overline 8 J.

8 0
3 years ago
Please help me! this is timed!
Arada [10]

Answer:

Radiation

Explanation:

Radiation refers to the emission of energy in rays or waves

8 0
3 years ago
A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9
ElenaW [278]

Answer:

The convection coefficient is 15456.48\ W/m^{2}K

Solution:

Mass flow rate, \dot{m} = 0.4\ kg

Inner diameter of the tube, d = 0.014 m

Fluid density, \rho_{f} = 990\ kg/m^{3}

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, P_{r} = 8.6

Heat flux, \dot{q} = 71,297\ W/m^{2}

Viscosity, \mu = 0.00079\ Ns/m^{2}

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}

Determine the velocity of the fluid inside the tube by mass flow rate:

\dot{m} = \rho_{f}A_{c}v

0.4 = 990\times 1.54\time 10^{- 4}v

v = 2.624 m/s

Determine the Reynold's Number, R_{e}:

R_{e} = \frac{\rho_{f}dv}{\mu}

R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253

Thus it is clear that R_{e} > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}

N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42

Also,

N_{u} = \frac{dh}{K}

where

h = convection coefficient

Now,

292.42 = \frac{0.014\times h}{0.74}

h = 15456.48\ W/m^{2}K

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brainly.com/question/11243732

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