(02.09 MC) An object starts at rest. Its acceleration over 30 seconds is shown in the graph below: An acceleration versus time g
2 answers:
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Answer:
The value is
Explanation:
From the question we are told that
The rate at which the plate rotates is
Generally the revolution per second is mathematically represented as
=>
Answer:
104.3 cm or 179.7
Explanation:
First find time that it takes for the object to hit the ground
*
Then find xf of projectile
not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7
hope that helps
Answer:
a) 80 V
b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge
Explanation:
<u>Given :</u>
We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( = 0) to the point B at distance l = 0.500m where its kinetic energy is
(
= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = +
30.0 v.
<u>Required :</u>
<em>(a) We are asked to find the electric potential VB </em>
<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>
<u> Solution </u>
(a) We are given the values for VA, and q so we want to find a relationship between these three parameters and VB to get the value of VB.
As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:
.........................................(1)
Where = 0 and the potential energy U of the charge is given by U = q V
where V is the electric potential. So, equation (1) will be in the form :
0+qVA= +qVB (Divide by q)
VA= /q + VB (solve for VB)
VB=VA- /q .......................................(2)
We get the relation between VB, VA and , now we can plug our values for VA,
and q into equation (2) to get VB
VB=VA- /q
=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)
=80 V
(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points
VA-VB =...................................(a)
VA-VB=E (solve for E)
E= VA-VB/..................................(3)
Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E
E= VA-VB/
=-100 N/C
The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge