What is the magnitude of the student's velocity after throwing the jacket? (absolute value)?
A 99 kg physics student is wearing a 5 kg jacket. He is stranded in the middle of a frictionless frozen pond. Since there is no friction, he cannot walk off the pond since his shoes can't grip the ice. Since he cannot escape the ice, surely he is doomed to freeze to death. Luckily, since the student is in the Lone Peak physics class, he knows conservation of momentum. He takes off his jacket and throws it east with a velocity of 17 m/s.
What is the magnitude of the net force?
A student who weighs 530 N is wearing a backpack that weighs 84 N. The student is standing still on level ground. Give your answers to the following questions in Newtons. a) What is the magnitude of the net force on the student? b)What is the magnitude of the contact force on the student by the backpack? c) What is the magnitude of the contact force on the student by the ground?
What is the magnitude and direction of the resultant force if two students are carrying a pail of water with 5N, and 19N at 30 degree respectively?
I'll explain how to get the answer, but I'm not going to just tell you.To determine resultant force you need to plot (x,y) points for both forces.We'll have one force at 0° and the other at 30°.To plot a point for each you need to use Cos and Sin.Multiply the magnitude of the force by Cos of the angle for “x" and repeat with Sin for “y".[math]x=5×Cos(0°) [/math][math]y=5×Sin(0°)[/math]x=5 y=0[math]x=19×Cos(30°)[/math][math]y=19×Sin(30°)[/math]x=16.45 y=9.5Then you add the two x’s and y's together for the final point. x=21.45 y=9.5Find Tan[math]^{-1}[/math] of y divided by x.Tan[math]^{-1}[/math](.44289)= 23.88°Resultant angle equals 23.88°.To determine the resultant magnitude you find the square root of x square plus y squared.[math]N=\sqrt{21.45^{2}+9.5^{2}}[/math]N[math]\approx[/math]23.46If you make the 19N 0° then the resultant angle is different, but the resultant magnitude is the same.
A student bus pass costs $11.25 for 15 rides. If each ride costs the same amount, which of the following represents the value v of the pass, in dollars, after r rides, where r is a positive integer less than or equal to 15?
Is this a trick question?11.25$/15Rides=.75 perRider=Number of RidesValuePass=.75r
What will be the magnitude of acceleration of A if the same force is acting on A and B and B's mass is 3 times that of A?
Let force on 'A' be 'F1' And force on 'B' be 'F2'.Similarly let their mass be 'M1' and 'M2'Given: F1 =F2, therefore let ' F’ be the force acting on them.And M2 = 3M1 ……..(1)Now let the accln be A1 nd A2 respectively.For AF= M1 A1 ……..(2)For BF= M2 A2 = 3M1 A2 ………(3)…….(from (1) )From (2) and (3).. we getM1 A1 = 3 M1 A2A1 = 3A2. I.e, the acceleration in A is 3 times the acceleration in B
A 59 kg student running at 5 m/s decelerates to a full stop. What is the impulse of the student?
295 N•s
Why is the national debt more of a concern than student loan debt?
They are dissimilar in that student loan debt has an identifyable individual who is accountable for paying the debt.But in addition to their education debt, these same people owe their part of the national debt and their part is about $400,000 on top of their student loan debt.But everybody owes the $400,000 so what happens with that? It means taxes will have to go higher and our standard of living will go lower as prices increase on everything.The mantra that a person with a college degree will make a million dollars more over their lifetime compared to someone without a degree is now a lie for most students. The students for whom that mantra still applies are in S.T.E.M. fields.
A student who weighs 480N is wearing a backpack that weighs 71N. The student is standing still on level ground?
a.)The sum of forces acting on the student to and from the bag is zero. However, the student will have a non-zero forces on the student's shoulder from the straps. Horizontal forces will be generated from the student's back towards the bag. Since there is no acceleration the sum of all these non-zero forces will be ZERO. To calculate all the non-zero forces we need more data such as strap angle and other geometrical data. b) As mentioned more geometrical data will be need to resolve these forces whose sum is zero. There are many contact forces in play, some are on the straps pulling the shoulder down. Other are reactive forces from the student's back. c) The reactive force from the ground is the sum of the student's weight and the bag. That is 551 N acting normally.
A student runs 30 m east, 40 m north, and 50 m west.?
So you go east 30m, then 40m north, then 50m west. When you go east you head in the +x direction 30m and when you go west, you go 50m in the -x direction, so the net x-displacement is: -50m + 30m = -20m--->The negative sign indicates 20m net displacement to the west. Your y-displacement is the 40m north, so now you you have a right triangle, with an unknown angle. You can use the Pythagorean theorem to solve for the magnitude of your resultant displacement vector: Resultant displacement = sqrt(20^2 + 40^2) To find the angle of your resultant vector use the inverse tangent function as such: inv tan(40m/20m) = Angle of your resultant vector above the x-axis. Hope this helps! :)