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Dahasolnce [82]
3 years ago
15

The weight of the block in the drawing is 97.0 N. The coefficient of static friction between the block and the vertical wall is

0.520. (a) What minimum magnitude of the force is required to prevent the block from sliding down the wall? (Hint: The static frictional force exerted by the block is directed upward, parallel to the wall.) (b) What minimum force is required to start the block moving up the wall? (Hint: The static frictional force is now directed down the wall.)
Physics
1 answer:
katrin [286]3 years ago
4 0

Hello,

I hope you're having a great day!

Here's what I got to question A and B

a) Fs(max)=(0.560)(88.9N)=49.8N Fy=88.9N-49.8N=39.1N Fx=(39.1N)/(cos(40))=51.0N sqrt{39.1^2+51.0^2}=F

F=64.3N

b) 88.9N +49.8 N=138.7N=Fy Fx=(138.7N)/(cos(40))=181.06N sqrt{138.7^2+181.06^2}=F

F=228N

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A 1,650 kg SUV comes uniformly to a stop. If the vehicle is accelerating at -1.3 m/s^2,
lions [1.4K]

Answer:b

Explanation:

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3 years ago
a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q
Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0   F=0, a=0

x= 0.050    F= -7.5 N,  a= -15 m/s²

x= 0.150     F= 22.5 N,  a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

          F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values ​​are from the equilibrium position of the spring.

Let's calculate the force

x = 0

      F = 0

x = 0.050

      F = - 150 0.050

      F = - 7.50 N

x = 0.150

      F = - 150 0.150

      F = - 22.5 N

let's use Newton's second law to find the acceleration

      F = m a

      a = F / m

x = 0 m

      a = 0

x = 0.050 m

      a = -7.50 / 0.50

      a = - 15 m / s²

x = 0.150 m

      a = - 22.5 / 0.50

      a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte  la fuerza es dada por la fuerza de Hoke,  

          F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es  1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza  

x=0  

              F= 0

x=0.050  

              F = - 150 0.050

              F= - 7.50 N

x= 0.150  

                F= - 150 0.150

                F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

          F = m a

          a = F/m

x =0  m

        a = 0

x= 0.050 m

         a = -7.50/ 0.50

          a =- 15 m/s²

x= 0.150 m

          a= - 22.5 / 0.50

          a= - 45 m/s²

7 0
3 years ago
If you place one staple on an electronic balance, the balance still reads 0.0 grams. However, if you place
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Mass of 1 staple = 6.8 g/210 staples
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Is friction and pushing similar ????
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Answer:

Yes, they are.

Explanation:

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Considering that we are talking about a stepdown transformer, and a turn ration of 1:24

Then

Vsecondary coil = 120 V / 24 = 5V

(But lets remember that the power must be conserved in the transformer, so the voltage is 24 times less, but the current is 24 times higher)

It provides 5 volts to operate the player or charge the batteries
8 0
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