The region where warm and cold air masses meet is called a front
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>
Answer:
Static energy
Explanation:
Think of it as a balloon rubbing against your hair, the two attractions of friction causes Static energy.
Answer:
Explanation:
To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.
To add the two vectors analytically you decompose each vector into their vertical and horizontal components.
<u>1. 18 mph 327º</u>
- Horizontal component: 18 mph × cos (327º) = 15.10 mph
- Vertical component: 18 mph × sin (327º) = - 9.80 mph
<u>2. 4 mph 60º</u>
- Horizontal component: 4 mph × cos (60º) = 2.00 mph
- Vertical component: 4 mph × sin (60º) = 3.46 mph
<u>3. Addition:</u>
You add the corresponding components:
To find the magnitude use Pythagorean theorem:
<u>4. Direction:</u>
Use the tangent ratio:
Find the inverse:
Answer:
I = 18 x 10⁻⁹ A = 18 nA
Explanation:
The current is defined as the flow of charge per unit time. Therefore,
I = q/t
where,
I = Average Current passing through nerve cell
q = Total flow of charges through nerve cell
t = time period of flow of charges
Here, in our case:
I = ?
q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C
t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s
Therefore,
I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)
<u>I = 18 x 10⁻⁹ A = 18 nA</u>
