Answer:
Explanation:
A. The ratio of their masses is always the same
Answer:
The velocity of the camera is 33.11 m/s.
Explanation:
Given that,
Speed = 10.8 m/s
Altitude = 50 m
Suppose determine the velocity of the camera just before it hits the ground?
We need to calculate the velocity of the camera
Using equation of motion
Where, v = final velocity of camera
u = initial speed of camera
s = distance
Put the value into the formula
The direction will be downward so it is the negative velocity.
Hence, The velocity of the camera is 33.11 m/s.
Answer:
E = 9.4 10⁶ N / C
, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
66 g of element Y is needed since the ratio between element X and element Y is 1:2
Answer:
Explanation:
The final velocity is given by the following kinematic equation:
Here, 
is the initial velocity, a is the body's acceleration and t is the motion time. We have to convert the time to seconds:
Now, we calculate the final velocity:
