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kifflom [539]
3 years ago
13

Current can also be called

Physics
2 answers:
dmitriy555 [2]3 years ago
6 0
D. It is electricity and moving charges
muminat3 years ago
4 0
Electrical current is the flow of moving charges.
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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
The amount of energy it takes to lift a box might be a function of which of the following?
Nesterboy [21]

The amount of energy it takes to lift a box might be a function of the weight of the objects inside the box. Work is proportional to force and distance. The force of the box is the weight itself of the box. Hence the answer to this problem is B.
5 0
3 years ago
Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speed up to
SIZIF [17.4K]

Answer:

5 m/s^2

Explanation:

The acceleration of the train is given by:

a=\frac{v-u}{t}

where:

u = 5 m/s is the initial velocity of the train

v = 45 m/s is the final velocity of the train

t = 8 s is the time taken for the train to increase the velocity

Substituting numbers into the formula, we find

a=\frac{45 m/s-5 m/s}{8 s}=5 m/s^2

5 0
4 years ago
Read 2 more answers
2. What is the percent composition of sulfur in H2SO4?
allochka39001 [22]

Answer:

C. 32.7%

Explanation:

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 % pls mark brainliest

6 0
3 years ago
The slope of the following graph represents
Harlamova29_29 [7]
Can you please provide an image of the graph?
5 0
3 years ago
Read 2 more answers
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