The correct answer to this question is <span>48.8N</span><span> on a bearing of </span><span>134.2 degrees.
In order to solve this, first we need to find the resultant force of the men pulling in the north and south directions:
F = 40 - 6 = 34N [due south - 180]
Then, find the resultant force.
R^2 =34^2 + 35^2
R^2 = 2381
R = sqrt(2381)
R = 44.8N
The angle from the vertical is given by:
tan(x) = 35/34
tan(x) = 1.0294
angle = 45.8 degrees
Taking N as zero degrees, then it bears 134.2 degrees.
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Answer:
Decelerating
Explanation:
One arrow is moving it forward but the other arrow is using more force to move it backwards causing it to have a reduction in speed.
Answer:
Explanation:
As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron
So here we have to find the work done by electric field on moving electron
So we have
now the distance moved by the electron is given as
so we have
now we have to convert it into keV units
so we have
Answer:
<em>17 m/s west</em>
Explanation:
Runner 1 has velocity = 10 m/s west
runner 2 has velocity = 7 m/s east
From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of
velocity = 10 m/s + 7 m/s = <em>17 m/s west</em>
Answer:
a) 5m/s2
b) 4 sec
c) 6 sec
d) 90 m
e) Answer in the file attached as it is a graph
Explanation:
This question can be solved using equations of motion. The two equations are:
2(a)(s) = v² – u²
v = u + (a)(t)
a = acceleration
s = distance
v = final velocity
u = initial velocity
t = time
a) Given final velocity = 50m/s and initial velocity = 30 m/s for 160m journey
Using 2(a)(s) = v² - u²
2(a)(160) = 50² – 30²
320 (a) = 2500 - 900
a = 1600/320
a = 5m/s²
b) The acceleration remains constant throughout so we can use it in this part as well.
Using v = u + (a)(t)
50 = 30 + (5)(t)
t = 4 sec
c) The acceleration remains constant throughout so we can use it in this part as well. Now the initial velocity will be 0m/s and final will be 30m/s
Using v = u + (a)(t)
30 = 0 + 5(t)
t = 6 sec
d) The acceleration remains constant throughout so we can use it in this part as well. Now the initial velocity will be 0m/s and final will be 30m/s
Using 2(a)(s) = v² - u²
2(5) (s) = 30² – 0²
10(s) = 900
s = 90 m
e) Graphs are attached as image