Answer:pressure = density * acceleration due to gravity * height
H=72.6cm= 0.726m
P=0.726*13.6*10^3*9.8
P=96761.28Pa
Explanation:
The force of gravity between the apple and the baseball is 3.335×10¯¹¹ N
The following data were obtained from the question given above:
Mass of apple (M₁) = 1 Kg
Mass of baseball (M₂) = 2 Kg
Distance apart (r) = 2 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
<h3>Force (F) =? </h3>
The force of gravity between them can be obtained as follow:
Therefore, the force of gravity between them is 3.335×10¯¹¹ N
Learn more on gravitational force: brainly.com/question/975812
Answer:
Explanation:
If magnetic field exists parallel to the direction of motion of electron , no force will act on the electrons and hence there will be no deflection in them .
In such cases , the magnitude of magnetic field is immaterial . No matter how high or low magnitude of magnetic field be , if it is parallel to the velocity of electron , it will not be deflected as the force created on them will be zero.
Only in case magnetic field makes some angle with the direction of velocity , force will be created and electron will be deflected .
Radius of the rings = 10 / 2 = 5 cm 0.05m
Distance between them D = 15 cm = 0.15 m
The charge of left ring is Ql = -20 x 10^-9 C
The charge of right ring is Qr = 20 x 10^-9 C
Since it the electric field at the midpoint, it is same for both rings and the there are opposite charges the electric filed will be sum of electric field of both charges.
Electric field on the left El = (k x Ql) / (D/2)^2 = (4 x k x Ql) / D^2
Electric field on the Right Er = (k x Qr) / (D/2)^2 = (4 x k x Qr) / D^2
Electric field E = El + Er = ((4 x k x Ql) / D^2) + ((4 x k x Qr) / D^2)
E = 4k/ D^2 (Ql + Qr); k is Coulombs constant = 8.99 x 10^9 Nm2Câ’2
E = 4 x 8.99 x 10^9/ 0.15^2 (2n + 2n) = 1598 x 10^9 x 4 x 10^-9 = 6393 N/c
Force f = E x q, q being the charge at the mid point
Force = q x 6393 N/c = 6.393q x 10^3 N
THAT LINK IS A VIRUS NEVER GO TO A LINK and if you go to “goggle” you should see a camera icon and take a picture of the question and get the answer there