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 = ..................1

put here value we will get

velocity in problems per hour =

**velocity in problems per hour = 4 per hour **

**so correct option is b. 4 per hour**

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.

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

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?

A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

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?

A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

**Answer:**

Volume of gasoline spills out is 0.943 L.

**Explanation:**

Volumetric expansion of both gasoline and steel tank is :

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We know expansion due to temperature change is :

For gasoline:

Similarly for Steel tank:

.

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

The **spacing **between the** two slits** is 0.221mm.

The **spacing** between the two slits is given as,

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

** Given **that,

**Substitute** in above equation.

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

Learn more about the** sodium lamp** here:

(b)how large was the acceleration, in units go g= 9.80 m/s ^2

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**

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)

**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

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

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

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

**Answer:**

**separation between the slits is 0.28 mm**

**Explanation:**

**given data **

wave length λ = 589 nm = 589 × m

distance between slits and the screen D = 0.91 m

fringes weight y = 0.19 cm = 0.19 × 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 × =

solve we get

**d = 0.28 mm**

**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**