Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
Answer:
13 km/h
Explanation:
Average speed = distance/time
Let the total distance and total time taken for the whole trip be d km and t hours respectively
Average speed for the whole trip = 82 km/h
d = 82t
The distance covered in the first half = d1/2
Time taken = t/2
Average speed = 69 km/h
69 = d1/2 ÷ t/2
d1 = 69t
The distance covered in the second half = d2/2
Time taken = t/2
Let the average sly for the see half be A
A = d2/2 ÷ t/2
d2 = At
d = d1 + d2
82t = 69t + At
At = 82t - 69t
At = 13t
A = 13t/t = 13 km/h
Answer:
Gamit ang graphic organizer na concept po punan niyo po ang mga sangay na ito at magpapakita ng paggalang sa kapwa kalaro
Answer:
The answer is below
Explanation:
The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm
radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m
a) Initial angular velocity (
) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s
θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
angular acceleration (α) is:
b)
c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
a) When it stops, the final angular velocity is 0. Hence:
θ = 323 rad
Answer:
8 N North.
Explanation:
Given that,
One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.
We need to find the magnitude of net force acting on the object.
Let North is positive and South is negative.
Net force,
F = 10 N +(-2 N)
= 8 N
So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".
