This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.
<h3>What is force?</h3>
Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Given data;
m(mass of sled)=8 kg
Θ is the inclination of force= 50°
Force of friction,f=2.4 N.
The applied force at the given angle is resolved into the two-component as;


The net vertical force is zero;

From Newton's second law the net force as;

Hence, the acceleration of the sled will be 1.30 m/s².
To learn more about the force refer to the link;
brainly.com/question/26115859
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The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m
Answer:
The objects morion will remain the same
Explanation:
Water travels through long, thin tubes running up from the roots through the stems and leaves called xylem.Water moves up the xylem through a process called capillary action.
Capillary action allows water to be pulled through the thin tubes because the molecules of the water are attracted to the molecules that make up the tube. The water molecules at the top are pulled up the tube and the water molecules below them are pulled along because of their attraction to the water molecules above them.