In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
Answer:
a) 
b)
c) 
d) Treat the humans as though they were points or uniform-density spheres.
Explanation:
Given:
- mass of Mars,

- radius of the Mars,

- mass of human,

a)
Gravitation force exerted by the Mars on the human body:

where:
= gravitational constant


b)
The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.


c)
When a similar person of the same mass is standing at a distance of 4 meters:


d)
The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.
- Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
- Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
Answer:
Option B (1.51 m)
Explanation:
U 2 can help me by marking as brainliest........
Answer:
41.4* 10^4 N.m^2/C
Explanation:
given:
E= 4.6 * 10^4 N/C
electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field
then electric flux = ∫ E*n dA
= ∫ 4.6 * 10^4 * 3*3
= 41.4* 10^4 N.m^2/C
Answer:
V = 20.5 m/s
Explanation:
Given,
The mass of the cart, m = 6 Kg
The initial speed of the cart, u = 4 m/s
The acceleration of the cart, a = 0.5 m/s²
The time interval of the cart, t = 30 s
The final velocity of the cart is given by the first equation of motion
v = u + at
= 4 + (0.5 x 30)
= 19 m/s
Hence the final velocity of cart at 30 seconds is, v = 19 m/s
The speed of the cart at the end of 3 seconds
V = 19 + (0.5 x 3)
= 20.5 m/s
Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s