Answer:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 ×
N
(d) 7.37 ×
N
Explanation:
(a) The minimum value of
will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

<em>Hence, the maximum distance is 5.002 m</em>
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force= 

×
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F = 
F= 7.37×
N
Answer:
2.7%
Explanation:
Given:
Uncertainty of the speedometer (u)= 2.5km/h
Speed measured at that uncertainty (v) = 92km/h
Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e
p =
%
p =
%
p = 2.7%
Therefore, the percent uncertainty is 2.7%
Answer:
An electron has a negative electric charge.
That is b. hope this helps cx
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know

where:
a= acceleration


we have:
initial velocity,
Using the eq. of motion:

where:
v= final velocity after the separation of spring with the bullet.


(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,

∵At maximum height the final velocity will be zero

Using the equation of motion:

where:
h= height
g= acceleration due to gravity


is the height from the release position of the spring.
So, the height from the latched position be:


