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Feliz [49]
3 years ago
9

Think about the pencil-dropping activity that you did in the introduction. What did the target finally look like?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer:

By dropping a pencil from a certain fixed height again and again it will make the target super messay with marks of dot everywhere on the target and some even out side the target.

Explanation:

You might be interested in
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
How much work, in N*m, is done when a 10.0 N force moves an object 2.5 m?
Alik [6]
W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
8 0
3 years ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
3 years ago
A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s
Taya2010 [7]
42.47 meters per second
5 0
3 years ago
If you run 12 m/s for 15 minutes, how far will you go?
vfiekz [6]
10800 m = 10.8 km should be the answer if I am correct
3 0
3 years ago
Read 2 more answers
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