Answer:
a
b
Explanation:
From the question we are told that
The diameter of the Ferris wheel is
The period of the Ferris wheel is
The mass of the passenger is
The apparent weight of the passenger at the lowest point is mathematically represented as
Where is the centripetal force on the passenger, which is mathematically represented as
Where is the angular velocity which is mathematically represented as
substituting values
and r is the radius which is evaluated as
substituting values
So
W is the weight which is mathematically represented as
So
The apparent weight of the passenger at the highest point is mathematically represented as
substituting values
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
2.12m/s is water and if it goes 1.5 seconds fast
I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.
If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.
I'm guessing that you're supposed to enter whatever the expression becomes
when either N₀ or ' t ' has some special value that's in the first line.
Just taking a wild guess here . . . . .
If it's "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is N₀ .
But that's just a wild guess. As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.
The calculated electric field is 3.0 N/C
<u>Explanation:</u>
The electric field is described only as the charge that it produces and is unique at every point in space. In particular, the electric field E is defined as the ratio of Coulomb forces to test charge. It can express as follows,
Where, F is the electrostatic (or Coulomb) force exerted on the positive test charge q. It is clear that E acts in the same direction of force. It is also assumed that q is so small and will not change the distribution of the charge generated by the electric field. It can express by a unit Newton per Coulomb (N / C).
Here, the given data, q = 2.0 C and F = 6.0 N. So, the electric field would be,