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Llana [10]
2 years ago
11

Alistar places a 6.4 kg block on an incline plane that is 23 degrees above the horizontal. The block starts from rest, and in 2.

2 seconds, travels 1.4 meters. What is the force of friction on the block?
Physics
1 answer:
crimeas [40]2 years ago
7 0

fh= 6.4×23

fh=147.2

fv=2.2×1.4

fv=3.08

f=147.2+3.08

f=150.28

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A student carried out an experiment adding different weights to a spring and recording the results. Look at the table of results
MAXImum [283]

Answer:

0.25 m.

Explanation:

We'll begin by calculating the spring constant of the spring.

From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:

Force (F) = 0.1 N

Extention (e) = 0.125 m

Spring constant (K) =?

F = Ke

0.1 = K x 0.125

Divide both side by 0.125

K = 0.1/0.125

K = 0.8 N/m

Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

Force (F) = 0.2 N

Spring constant (K) = 0.8 N/m

Extention (e) =..?

F = Ke

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Divide both side by 0.8

e = 0.2/0.8

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5 0
3 years ago
Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
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Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

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