Answer:
Time period of oscillations is 0.62 s
Explanation:
Due to suspension of weight the change in the length of the spring is given as
now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force
Now the period of oscillation of spring is given as
Now plug in all values in it
Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :
F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
Jennifer runs 5 miles east, then stops to take a break. after her break, she continues running 4 more miles east.
In this the total distance she covers is 9 miles and the total displacement is also the same as the distance that is 9 miles.
<h3>What is the difference between distance and displacement?</h3>
Displacement is the shortest distance between initial and final position, or we can say it is the straight-line distance between initial and final position.
Whereas distance is considered as the total path length covered from initial position till the final position. The Displacement of a body is always less than or equal to the distance.
Displacement can be zero in case the initial and final positions coincide, but distance can never be zero.
To know more about distance and displacement, visit:
brainly.com/question/3243551
#SPJ4
Answer:
El microscopio y el descubrimiento de microorganismos. Antonie van Leeuwenhoek (1632-1723) fue una de las primeras personas en observar microorganismos, utilizando un microscopio de su propio diseño, e hizo una de las contribuciones más importantes a la biología. Robert Hooke fue el primero en usar un microscopio para observar seres vivos.
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
