A cat runs 80 meters to the left in 16 seconds, then 50 meters to the right in 10 seconds, then 70 meters to the left in 14 seco
1 answer:
<h3>Answer:a) Distance= 240m. b) Displacement= 90m. c) Average Speed= 5
. d) Average Velocity= 1.875 </h3>
Explanation:
a) Distance ➤ (80+50+70+40) m. ➤ Distance= 240m.
b) Displacement ➤ (50+40)m. ➤ Displacement= 90m
c) Average Speed ➤ Total distance travelled/total time taken
➤ 240/48
➤ 5
d) Average Velocity ➤ Total displacement/total time taken
➤ 90/48
➤ 1.875
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Answer:
171.5 N
Explanation:
The gravitational force on an object due to the Earth is given by
where
m is the mass of the object
g is the acceleration due to gravity
The acceleration due to gravity at a certain height h above the Earth is given by
where:
G is the gravitational constant
is the Earth's mass
is the Earth's radius
Here,
So the acceleration due to gravity is
We know that the mass of the object is
m = 70 kg
So, the gravitational force on it is
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: