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Montano1993 [528]
2 years ago
6

A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied

along string A at the point (5,8). Calculate the acceleration of the collar in the direction of the beam.
Physics
1 answer:
faltersainse [42]2 years ago
3 0

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

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Determina el tamaño de la arista de un cubo hecho de plata, cuya masa es de 10.49 kg.
Sidana [21]

Answer:

88.2 N

Explanation:

Datos

Lcubo = 10 cm = 0.1 m

Vcubo = Vfluido desalojado= 0.1 m x 0.1 m x 0.1 m = 10-3 m

mcubo = 10 kg

dfluido = 1000 kg/m3

g = 9.8 m/s2  

Sabemos que el peso aparente de un cuerpo que se sumerge en un fluido es:

Paparente=Preal−Pfluido

Teniendo en cuenta que:

Preal = mcubo⋅gPfluido=E= dfluido⋅Vfluido⋅g

Como el cuerpo se sumerge completamente en el fluido, el volumen de fluido desalojado es exactamente el volumen del cubo. Por lo tanto si sustituimos los datos que nos proporcionan en el enunciado en la primera ecuación:

Paparente=mcubo⋅g−dfluido⋅Vfluido⋅g ⇒Paparente=10 kg ⋅9.8 m/s2 − 1000 kg/m3 ⋅10−3 m ⋅9.8 m/s2 ⇒Paparente = 88.2 N

7 0
2 years ago
After scientists analyze the results of their experiments they
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They communicate their result to the scientific community- so to speak
8 0
3 years ago
9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

5 0
3 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
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Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

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2 years ago
A person walks a distance of 3.0 km due
Nataly [62]

Answer:

Speed=1.6km/hr. I'm not sure about b

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