For part 2 number 1:
first multiply 3.0 by 20 to get 60. Then multiply 3.0 y 10 to get 30. Add both answers together to get 90. The answer is 90 meters per second squared.
Answer:
Option D- 1,2 and 3
Explanation:
Gravity always acts on an object irrespective of height
Answer:
<em>b - wires</em>
Explanation:
cells , dry cells , electrical outlets are all responsible for the varying electron pressure/ potential difference hence wires is our answer because wires are just conductors which can only influence resistance.
Answer:
A horizontal line on a speed-time graph represents a constant speed.
Explanation:
Acceleration is defined by a sloping line on a speed-time graph. The sloping line shows the object's speed is increasing. The target either accelerates or slows down.
Velocity-time graphs are being used to illustrate the acceleration for moving objects in a horizontal path. You can use them to display momentum and to sort out movement.
Answer:
Supongo que queremos determinar la carga de la carga de prueba.
Sabemos que la fuerza culombiana entre dos cargas q₁ y q₂, separadas por una distancia R, está dada por:
![F = k_c*\frac{|q_1*q_2|}{R^2}](https://tex.z-dn.net/?f=F%20%3D%20k_c%2A%5Cfrac%7B%7Cq_1%2Aq_2%7C%7D%7BR%5E2%7D)
Aislandolo para una de las cargas, obtenemos:
![\frac{F*R^2}{kc*|q_1|} =|q_2|](https://tex.z-dn.net/?f=%5Cfrac%7BF%2AR%5E2%7D%7Bkc%2A%7Cq_1%7C%7D%20%3D%7Cq_2%7C)
En este caso sabemos:
fuerza atractiva, por lo que los signos de las cargas son opuestos.
q₁ = 5 mC
R = 35cm
F = 1.5 N
Kc = 9*10^9 N*m^2/C^2
Un primer paso, seria reescribir todos los valores en las mismas unidades.
Sabiendo que:
100cm = 1m
R = 35cm = (35/100) m = 0.35m
Y sabiendo que:
1mc = 1*10^(-6) C
Entonces:
q₁ = 5 mC = 5*1*10^(-6) C = 5*10^(-6) C
Ahora podemos reemplazar esos valores en la ecuación de la fuerza, para obtener el valor de la otra carga:
![\frac{15N*(0.35m)^2}{(9*10^9 N*m^2/C^2)*5*10^{-6}C} =|q_2| = 4.08*10^{-5} C](https://tex.z-dn.net/?f=%5Cfrac%7B15N%2A%280.35m%29%5E2%7D%7B%289%2A10%5E9%20N%2Am%5E2%2FC%5E2%29%2A5%2A10%5E%7B-6%7DC%7D%20%3D%7Cq_2%7C%20%3D%204.08%2A10%5E%7B-5%7D%20C)
Y recordar que las cargas tienen signo opuesto, entonces la carga de la carga de prueba es:
q₂ = -4.08*10^-5 C