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2 years ago

En la Tierra, una balanza muestra que tu peso es 585 N.

Physics

1 answer:

ra1l [238]2 years ago

5 0

Answer:

a) m = 59.63 [kg]

b) Wm = 95.41 [N]

Explanation:

El peso de un cuerpo se define como el producto de la masa por la aceleración gravitacional. DE esta manera tenemos:

W = m*g

Donde:

m = masa [kg]

g = gravedad = 9.81 [m/s^2]

m = W / g

m = 585 / 9.81

m = 59.63 [kg]

Es importante aclarar que la masa se conserva independientemente de la ubicación del cuerpo en el espacio.

Por ende su masa sera la misma en la luna.

El peso en la luna se calcula como Wm y es igual a:

Wm = 59.63 * 1.6 = 95.41 [N]

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Answer:

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Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

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So, f = 17.5 Hz

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2 years ago

Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-

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Answer:

a) W_total = 8240 J , b) W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

F - W = m a

W = mg

F = m (a + g)

F = 80 (1 + 9.8)

F = 864 N

The work of this force we will call it W1

We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

F₂ - W = 0

F₂ = W

F₂ = 80 9.8

F₂ = 784 N

The work of this fura we will call them W2

The total work is

W_total = W₁ + W₂

W_total = (F + F₂) y

W_total = (864 + 784) 5

W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

W₁ / W₂ = F y / F₂ y

W₁ / W₂ = 864/784

W₁ / W₂ = 1.1

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2 years ago

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

gayaneshka [121]

Answer:

Explanation:

Solution is in the picture attached

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2 years ago

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

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Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

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2 years ago