ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach
ed at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick
1 answer:
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
You might be interested in
Answer:
The speed of the package of mass m right before the collision
Their common speed after the collision
Height achieved by the package of mass m when it rebounds
Explanation:
Have a look to the diagrams attached below.
a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.
where is Kinetic energy and
is Potential energy.
and
Considering the fact we will plug out he values of the given terms.
So
Keypoints:
- Sum of energies and momentum are conserved in all collisions.
- Sum of KE and PE is also known as Mechanical energy.
- Only KE is conserved for elastic collision.
- for elastic collison we have
that is co-efficient of restitution.
<u>KE = Kinetic Energy and PE = Potential Energy</u>
b.Now when the package stick together there momentum is conserved.
Using law of conservation of momentum.
where
.
Plugging the values we have
Cancelling m from both sides and dividing 3 on both sides.
Law of conservation of energy will be followed over here.
c.Now the collision is perfectly elastic
We have to find the value of for m mass.
As here we can use that if both are moving in right ward with
then there is a
velocity when they have to move leftward.
The best option is to use the formulas given in third slide to calculate final velocity of object .
So
Now using law of conservation of energy.
The linear momentum is conserved before and after this perfectly elastic collision.
So for part a we have the speed for part b we have their common speed
and for part c we have the rebound height
.
Answer:
a) t = 3.3 [h]
b ) Hora = 13:18 o 1:18 [pm]
c) x = 327.79 [km/h] (Elena)
x = 196.6 [km] (Javier)
Explanation:
Para poder solucionar este problema debemos hacer un planteamiento inicial de ubicacion de las ciudades, este planteamiento nos ayudara a entender el problema de una manera mas facil.
Tenemos las ciudades A & B y la ciudad de Madrid que esta a una distancia x con respecto de B, (ver esquema adjunto).
De manera logica debemos deducir que la Ciudad A debe estar mas lejos de Madrid que la ciudad B de la misma Madrid, ya que en caso contrario Javier nunca alcanzara a Elena, ya que Elena va mas rapido que Javier.
a) Ahora debemos de utilizar la siguiene ecuacion de la cinematica, cuando los cuerpos se mueven a velocidad constante.
Donde:
x -xo = Distancia entre el punto inicial y punto final.
v = velocidad [m/s]
t = tiempo [s]
Debemos convertir las velocidades de kilometros por hora a metros por segundo.
Seguidamente formulamos una ecuacion por cada movimiento, luego debemos igualar estas ecuaciones en funcion de la variable x que sera el punto donde se encuentren ambas personas.
<u>Para Elena</u>
<u>Para Javier</u>
<u /><u />
Igualamos las variables x de ambas ecuaciones.
b) La hora facilmente se puede encontrar sumando el tiempo con las 10:00am
hora = 10 + 3 = 13 [hrs]
La parte decimal debe convertirse a tiempo.
Hora = 13:18 o 1:18 [pm]
c) Para encontrar estas distancias utilizamos el tiempo encontrado en el item a.
<u>Para Elena</u>
<u>Para Javier</u>
<u /><u />
