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Dafna1 [17]
3 years ago
14

(3.16_Q2) Which weights would you use on a single thread to create a 6.86 N force? Question 2 options: Weight IDs A, B, C, D Wei

ght IDs D, E, F Weight IDs E, F Weight IDs B, C, E, G Question 3 (1 point) (3.16_Q3) Which weights would you use on a single thread to create a 7.84 N force? Question 3 options: Weight IDs A, B, C, D Weight IDs D, E, F Weight IDs E, F Weight IDs B, C, E, G Question 4 (1 point) (3.16_Q4) Which weights would you use on a single thread to create a 12.45 N force? Question 4 options: Weight IDs A, B, C, D Weight IDs D, E, F Weight IDs E, F Weight IDs B, C, E, G Question 5 (1 point) (3.16_Q5) Which weights would you use on a single thread to create a 1.76 N force? Question 5 options: Weight IDs A, B, C, D Weight IDs D, E, F Weight IDs E, F Weight IDs B, C, E, G 0 of 5 questions saved
Physics
1 answer:
Tomtit [17]3 years ago
5 0

1. E,F

2. D,E,F

3. B,C,E,G

4. A,B,C,D

I did the test! :)

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Two concentric current loops lie in the same plane. The smaller loop has a radius of 2.7cm and a current of 12 A. The bigger loo
Kryger [21]

Answer:

Radius of bigger loop(R) =4.5cm

Explanation:

Consider a circular path of radius r around the wire.  The magnetic field along that path is given by ;

∫B*dl = k*I where I is the current enclosed.  From symmetry, ∫B*dl = 2*π*r*B

B = K*I/r, so the magnetic field varies inversely as the loop radius and directly as the current.

The smaller loop current to radius ratio is  12/2.7

The bigger loop current to radius ratio is = 20/R

12/2.7 = 20/R

R = (20 * 2.7)/12

R=54/12

R=4.5cm

8 0
3 years ago
Read 2 more answers
What is the fundamental property of matter that can either be positive or negative?
larisa86 [58]
Protons, electrons, and neutrons have electrical charges. Protons are positive. Electrons are negative and neutrons are neutral (no charge).<span />
5 0
3 years ago
En un lago helado se lanza un trozo de hielo de 500 g a la velocidad de 20 m/s. Si el coeficiente de rozamiento es de 0.04, calc
nika2105 [10]

Answer:

Sabemos que:

La masa del trozo de hielo es 500g

la velocidad inicial es 20 m/s

el coeficiente de fricción es 0.04

a) La fuerza de rozamiento de un objeto de masa M es escrita como:

F = M*g*μ

Y es en dirección opuesta al movimiento del objeto, entonces la rescribimos como:

F = -M*g*μ

Donde M es la masa del objeto, 500g, es útil escribirla en kilogramos, entonces podemos escribir M = 500g = 0.5 kg

g es la aceleración de la gravedad:

g = 9.8m/s^2

y μ es el coeficiente de fricción, en este caso es 0.04

Entonces la fuerza de fricción es:

F = -0.5kg*9.8m/s^2*0.04 = -0.196 N

b) Ahora queremos la aceleración, tenemos la segunda ley de Newton que dice:

F = M*a

-0.196 N = 0.5kg*a

(-0.196 N)/0.5kg = a = -0.392 m/s^2

c) Primero debemos escribir la ecuación de movimiento.

La aceleración es:

a = -0.392 m/s^2

Para obtener la velocidad, debemos integrar sobre el tiempo para obtener.

v(t) = (-0.392 m/s^2)*t + v0

Donde v0 es la velocidad inicial, en este caso 20m/s

Entonces la ecuación es:

v(t) = (-0.392 m/s^2)*t + 20m/s

Recordemos que esta fuerza solo actua mientras el objeto se mueva, esto significa que cuando la velocidad sea igual a cero, la fuerza desaparece, por lo tanto el objeto se detiene.

Entonces podemos obtener el valor de t para el cual la velocidad es igual a cero.

v(t) = 0 = (-0.392 m/s^2)*t + 20m/s

(0.392 m/s^2)*t  = 20m/s

t = 20m/s/(0.392 m/s^2) = 51.02 s

Esto significa que el objeto se va a mover por 51.02 segundos antes de detenerse totalmente.

Ahora, la ecuación de la posición puede obtenerse si integramos la ecuación de la velocidad, así obtenemos:

p(t) =  (1/2)*(-0.392 m/s^2)*t^2 + 20m/s*t + p0

donde p0 es la posición inicial del objeto.

Ahora, el espacio total recorrido por el trozo de hielo va a ser igual a la diferencia entre la posición final y la posición inicial, esto es:

D = p(51.02s) - p(0s)

D = ((1/2)*(-0.392 m/s^2)*(51.02s)^2 + 20m/s*51.02s + p0) - ((1/2)*(-0.392 m/s^2)*(0s)^2 + 20m/s*(0s) + p0)

D = (1/2)*(-0.392 m/s^2)*(51.02s)^2 + 20m/s*51.02s = 510.20 m

El trozo de hielo se mueve por 510.20 metros.

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what causes the ocean waves that surfers use for their sport? a. density differences in water b. salinity differences in water c
irinina [24]
D) wind blowing across the water
3 0
3 years ago
Read 2 more answers
A force is exerted on a charge carrying particle by a ________ _____ only when the particle moves ________the field lines. If a
VMariaS [17]

Answer:

  force is exerted on a charge carrying particle by a MAGNETIC FIELD only when the particle moves PERPENDICULAR the field lines. If a charged particle moves parallel to the magnetic field, the force exerted by the magnetic field on the charged particle is_ ZERO

Explanation:

This complementation exercise refers to the magnetic force that the formula has

         F = q v x B

where the bold letters indicate vents and the x the vector product. The module of this product is

         F = qv B sin θ

where θ  is the angle between the speed and direction of the magnetic field.

It can be seen that for the force to be different from zero, the angle must not be parallel θ  = 0

        sin 0 = 0

the maximum value is obtained for an angle of θ  = 90º

       sin 90 = 1

Using these concepts we see that the missing words are.

  force is exerted on a charge carrying particle by a MAGNETIC FIELD only when the particle moves PERPENDICULAR the field lines. If a charged particle moves parallel to the magnetic field, the force exerted by the magnetic field on the charged particle is_ ZERO_______

5 0
2 years ago
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