Answer:
25. Approximately 8.1 meters
26. North 1.31 km, and East 2.81 km
Explanation:
25.
Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:
which can be rounded to 8.1 m.
26.
Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.
We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):
and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.
Citric Acid is the correct answer because it contains a density of 1.66 g/cm3, whereas water= 1.00 g/cm3, Olive oil= 0.93 g/cm3, Ethyl alcohol= 0.79 g/cm3
Answer:
Usually the coefficient of friction remains unchanged
Explanation:
The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.
Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s
