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

Answers

Answer 1
Answer:

Answer:

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Explanation:

given data

worked on  homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour

solution

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
Answer:

Answer:

option (b) is correct

Explanation:

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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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?

Answers

Answer:

Volume of gasoline spills out is 0.943 L.

Explanation:

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?

Answers

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:

brainly.com/question/24867424

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

Answers

acceleration = (velocity final-velocity initial)/ time
where
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)?

Answers

Answer:

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

Explanation:

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^2=u^2+2as

where

'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

v=u+at

where symbols have the usual meaning

Applying the given values we get

t=(v-u)/(g)\n\nt=(0-1.3966)/(-9.81)=0.1423seconds

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.

Answers

Answer:

separation between the slits is 0.28 mm

Explanation:

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

solution

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?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

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

Computation:

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