Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. What is the velocity in problems per hour? a. 6 per hour
b. 4 per hour
c. 10 per hour
d. 0.67 per hour
e. 15 per hour


Answer 1


velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour


given data

worked on  homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour


we know that Shirley solve complete 6 accounting homework problem in 1.5 hour so her velocity  in problems per hour will be as

velocity in problems per hour =  (complete\ problem)/(time\ taken)   ..................1

put here value we will get

velocity in problems per hour =  (6)/(1.5)

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Answer 2


option (b) is correct


time t = 1 and half hour = 1.5 hour

Number of problems, n = 6

So, the velocity in problems

v = n / t

v = 6 / 1.5

v = 4

So, the rate is 4 problems per hour.

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As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?
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Early in the morning, when the temperature is 5.5 °C, gasoline is pumped into a car’s 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 °C. Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out in this case?



Volume of gasoline spills out is 0.943 L.


Volumetric expansion of both gasoline and steel tank is :

\beta_(gas)=9.5 *10^(-4)/K\n\beta_(steel \ gas)=3.6 * 10^(-5)/K.  { source Internet}

We know expansion due to temperature change is :

\Delta V=\beta*\Delta T* V

For gasoline:

\Delta V_g=0.98 \ L.\n

Similarly for Steel tank:

\Delta V_(steel \ gas)=0.037\ L.

Now, volume of gasoline spills out is equal to difference between expansion in volume.

\Delta V_(gas)-\Delta V_(Steel \ gas)=0.98-0.037\ L=0.943\ L.

Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?


The spacing between the two slits is 0.221mm.

The spacing  between the two slits is given as,

                    D=(\lambda L)/(y)

Where \lambda is wavelength, y is fringe spacing and L is length of screen.

Given that, \lambda=589nm,L=150cm,y=4mm

Substitute in above equation.

               D=(589*10^(-9)*150*10^(-2)  )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm

Hence, the spacing between the two slits is 0.221mm.

Learn more about the sodium lamp here:

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2


acceleration = (velocity final-velocity initial)/ time
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
                     = 37.5 m/s
velocity initial = 35 km/hr x  1hr /3600 s x 1000 m/1 km
                      =  9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units  = (2.646m/s^2)/(9.8m/s^2)
                                             0.27 units

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?



The time for final 15 cm of the jump equals 0.1423 seconds.


The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as



'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

0^2=u^2-2* 9.81* 0.76\n\n\therefore u=√(2* 9.81* 0.76)=3.86m/s

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

v^(2)=3.86^2-2* 9.81* 0.66\n\n\therefore v=√(3.86^2-2* 9.81* 0.66)=1.3966m/s

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as


where symbols have the usual meaning

Applying the given values we get


Light from a sodium vapor lamp (λ-589 nm) forms an interference pattern on a screen 0.91 m from a pair of slits in a double-slit experiment. The bright fringes near the center of the pattern are 0.19 cm apart. Determine the separation between the slits. Assume the small-angle approximation is valid here.



separation between the slits is 0.28 mm


given data

wave length λ = 589 nm = 589 × 10^(-9) m

distance between slits and the screen D = 0.91 m

fringes weight y = 0.19 cm = 0.19 × 10^(-2) m


we find here the spacing between the two slits i.e d

so use here formula that is

y = λD ÷ d       .........................1

put here value we get

0.19 × 10^(-2) = (589*10^(-9)*0.91)/(d)

solve we get

d = 0.28 mm

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?



Time taken for 1 swing = 3.81 second



Time taken for 1 swing = 2.20 Sec


Time taken for 1 swing , when triple the length(T2)


Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second