By how much does the temperature of mercury increase? the specific heat of mercury is 140

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

An object weighs 40n in air, weighs 20n when submerged in water and 30n when submerged in a liquid of unknown density. what is the density of the liquid?

(2) (a) you have a simple circuit that consists of only a battery (δvbat=1.5v) and two resistors with resistance r1=10ω and r2=5ω, connected in series with each other. what is the current running through the battery? (b) you re-arrange your circuit so now r2 is attached in parallel to r1 rather than in series. what is the current running through the battery? (c) you add an additional resistor r3=7ω on the same branch as r2. what is the current running through the battery? (d) what is the power dissipated in r3?