Text book: We can measure the mass of the text book easily by weighing machine, to measure the volume we need to measure the length, width, and height of the text book by the ruler, by multiplying these dimension we can get the volume of the text book, and by dividing the mass of the book with its volume we can get the density of the book.
Milk Container: We can measure the mass of the milk container easily by weighing machine, now (assuming the milk container is cylindrical in shape) we need to measure its height, and and diameter and by the formula (π*r^2*h) we can measure its volume, and and by dividing the mass with its volume we can get the density of the milk container.
Air filled balloon: we can measure the mass of the air filled balloon by weighing it weight machine, we know that the density of air is 28.97 kg/m^3, by dividing the mass of the balloon with the denisty of air we can get the volume of the balloon.
K = 1/2 m x v^2
m = mass on the cart
V = velocity imparted to the cart
KA = 1/2 mA x vA^2.......................(1)
KB = 1/2 mB x vB^2........................(2)
Diving equation 1 by equation 2, we get -
KA/KB = mA/mB
= 2
KA = 2 x KB
Option A is correct
The water molecules would slow down, and as they slow down, the heat created from their movement would cease.
Measure a whole stack (one in which you know the number of sheets), then divide your measurement by the number of sheets in that stack
Answer:
a) Ws = 2.548 J
b) Wf = 1.153 J
c) v = 1.923 m / s
Explanation:
a) The work done by the spring force
Ws = ½ * k * x²
Ws = ½ * 260 N/m *0.14² m
Ws = 2.548J
b) The increase in thermal energy can by find using
Et = Wf
Wf = µ * m *g * x
Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m
Wf = 1.153 J
c) The speed just as the block reaches can by fin using
EK = Ws + Et
Ek = ( 2.548 + 1.153 ) J = 3.7 J
Ek = ½ * m * v²
v² = 2* Ek / m
v = √[2 * 3.7 J / 2.0 kg]
v = 1.923 m / s
