Answer:
q=1.4*10^{-9}C
Explanation:
Given data:
charge on ruler = -14μC
Mass of tissue is 5 g
To Know the minimum charge, equate electrostatic force to weight
we have F = W
so
putting all value in equation,

solving for q

or q=1.4*10^{-9}C
The linear speed of the ladybug is 4.1 m/s
Explanation:
First of all, we need to find the angular speed of the lady bug. This is given by:

where
T is the period of revolution
The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:
T = 1 s
Therefore, the angular speed is

Now we can find the linear speed of the ladybug, which is given by

where:
is the angular speed
r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation
Substituting, we find

Learn more about angular speed:
brainly.com/question/9575487
brainly.com/question/9329700
brainly.com/question/2506028
#LearnwithBrainly
Answer:
C- what is determined as right or wrong is universally defined and agreed upon.
Explanation:
Correct on edge
Ocean breezes are due to the convection method of heat transfer
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
#SPJ1