Answer:
a) 141.6m
b) 8.4m/s
Explanation:
a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:
hence, the total distance is 141.6m
b) the mean velocity of the total trajectory is given by:
hence, the mean velocity is 8.4 m/s
Answer:
question5: F=74312.5N
question6: charge at the end of antenna=0.37N
Explanation:
Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.
⇒
∴
where is the force of attraction or repulsion
is Coulumb's constant=
and
are the magnitude of the charges
is the distance between two charges
The force between the two charges is attractive if they are of different polarity
The force between the two charges is repulsive if they are of same polarity
Question5:
Given: q1=0.041 C, q2=0.029 C, r=12 m
therefore by Coulumb's law,
Question6:
Given: q1=, r=5 m, F=
therefore by Coulumb's law,
⇒
Since Astronaut and wrench system is isolated in the space and there is no external force on it
So here momentum of the system will remain conserved
so here we can say
initially both are at rest
so here plug in all values
so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s