Answer:
trial and error.
Explanation:
a problem solving method that involves trying all possible solutions until one works is using trail and error.
Explanation:
what is asked give brief explanation about question
26.54 m/s is the magnitude of its velocity just before it strikes the ground
h=100m,v=20m/s,g=9.8m/s
time it takes to reach the ground,
![[t=\sqrt2h/g],[=\sqrt2*100/9.8=4.51s]](https://tex.z-dn.net/?f=%5Bt%3D%5Csqrt2h%2Fg%5D%2C%5B%3D%5Csqrt2%2A100%2F9.8%3D4.51s%5D)
x= 120m
t= 4.52
v= x/t
v= 120/4.52
v= 26.54 m/s
The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.
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The coefficient of friction must be 0.196
Explanation:
For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:
where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:
is the coefficient of friction between the tires and the road
m is the mass of the car
is the acceleration of gravity
v is the speed of the car
r is the radius of the curve
In this problem,
r = 750 m is the radius
is the speed
And solving for
, we find the coefficient of friction required to keep the car in circular motion:

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Answer:

Explanation:
The strength of the electric field produced by a charge Q is given by

where
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is

and the fish can detect the electric field at a distance of

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:
