Calculate the speed of a bike rider who accelerates (from rest) for 5 seconds down a hill at an acceleration of 8
1 answer:
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Answer:
<u>20 Minutes</u>
<u></u>
Explanation:
Well we know Mph (Miles per hour) is distance over time :
R (rate) = 60
d (distance) = 20
t (time) = Unknown
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
R =
↓
60 =
↓
t =
↓
t = or 0.3333
<em>So basically it would take one third of an hour. Lets change these units to minutes.</em>
60 * 0.333333 = 20
<em>So it would take you </em><u><em>20 minutes</em></u><em> to drive 20 miles on a bus that drives 60 mph</em>
<em />
Hope that helps
<em>~Siascon~</em>
Answer:
Figure a. E_net = 99.518 N/C
Figure b. E_net = 177.151 N / C
Explanation:
Given:
- Attachment for figures missing in the question.
- The dimensions for rectangle are = 7.79 x 3.99 cm
- All four charges have equal magnitude Q = 10.6*10^-12 C
Find:
Find the magnitude of the electric field at the center of the rectangle in Figures a and b.
Solution:
- The Electric field generated by an charged particle Q at a distance r is given by:
E = k*Q / r^2
- Where, k is the coulomb's constant = 8.99 * 10^9
Part a)
- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:
E_1 + E_3 = 0
- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:
E_net = E_2 + E_4
E_2 = E_4
E_net = 2*E = 2*k*Q / r^2
- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:
r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )
r = sqrt ( 1.9151*10^-3 ) = 0.043762 m
- Plug the values in the E_net expression developed above:
E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3
E_net = 99.518 N/C
Part b)
- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).
- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.
Hence,
E_net = 2*E_part(a)*cos(Q)
- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:
Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.
- Now, compute the net electric field E_net:
E_net = 2*(99.518)*cos(27.12)
E_net = 177.151 N / C
