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Andru [333]
3 years ago
8

consider a charge of -15.0 mCmoving to the right a 2.00x10^6 m/s in a mganetic field of .0300 T pointing upwards. What is the ma

gitude and direction of the force on this charge? If this charge had a mass of 8.00 x10^10kg what would be the acceleration be
Physics
1 answer:
77julia77 [94]3 years ago
5 0

Answer:

a = 1.13×10^-8 m/s²

Explanation:

f = ma = qvb \\ a =  \frac{qvb}{m}

q = 15×10^-3 C

v = 2×10^6 m/s

B = 0.03 T

m = 8.00 x10^10 kg

a = 900/8 x10^10

a = 1.13×10^-8 m/s²

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Steroid use in females can cause masculine characteristics, such as lower voice and increased body hair." True False
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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

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Strike441 [17]
True because well it’s moving fast lol sometimes ur eyes have a hard time following its speed
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3 years ago
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