Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a band to make this much of a change in pitch due to the Doppler effect


Answer 1


-3.396 m/s or 3.465 m/s


v = Speed of sound in air = 343 m/s

v_s = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean f=1.01f'

From the Doppler effect equation we have

f=f'(v)/(v+v_s)\n\Rightarrow 1.01f'=f'(v)/(v+v_s)\n\Rightarrow 1.01=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(1.01)-343\n\Rightarrow v_s=-3.396\ m/s

The velocity is -3.396 m/s

when f=0.99f'

f=f'(v)/(v+v_s)\n\Rightarrow 0.99f'=f'(v)/(v+v_s)\n\Rightarrow 0.99=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(0.99)-343\n\Rightarrow v_s=3.46464646465\ m/s

The velocity is 3.465 m/s

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Which one of these are true about scale models? a. A map may have a scale model with the proportion of one centimeter to one kilometer. b. A scale model is always smaller than the object it represents, c. A model may be built to different scales.



Option A is true


For option A, it's true because a map that has a scale model with the proportion of one centimeter to one kilometer is known as verbal scale which is a type of scale.

For option B, it's not true because though a scale model is most times always smaller than the object it represents, there are sometimes when the scale model is an enlarged view/representation of a small object.

For option C, it's true because there is no one scale that is confined to just a model. A model can use different scales to depict an object.

A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her



The average upward force exerted by the water is 988.2 N



mass of the diver, m = 60 kg

height of the board above the water, h = 10 m

time when her feet touched the water, t = 2.10 s

The final velocity of the diver, when she is under the influence of acceleration of free  fall.

V² = U² + 2gh


V is the final velocity

U is the initial velocity = 0

g is acceleration due gravity

h is the height of fall

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 10

V² = 196

V = √196

V = 14 m/s

Acceleration of the diver during 2.10 s before her feet touched the water.

14 m/s is her initial velocity at this sage,

her final velocity at this stage is zero (0)

V = U + at

0 = 14 + 2.1(a)

2.1a = -14

a = -14 / 2.1

a = -6.67 m/s²

The average upward force exerted by the water;

F_(on\ diver) = mg - F_( \ water)\n\nma = mg - F_( \ water)\n\nF_( \ water) = mg - ma\n\nF_( \ water) = m(g-a)\n\nF_( \ water) = 60[9.8-(-6.67)]\n\nF_( \ water) = 60 (9.8+6.67)\n\nF_( \ water) = 60(16.47)\n\nF_( \ water) = 988.2 \ N

Therefore, the average upward force exerted by the water is 988.2 N

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?


A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = (2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) )  = (106x)/(x+53)

                              26.5 = (106x)/(x+53) \n \n 79.5 x = 1404.5\n \n x = 17.67 miles/hour

Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , measured relative to the merry-go-round. neglect friction and the size of each child.


by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

m*2*R + Iw = 0

w = -(2mR)/(I)

so merry go round will turn in opposite direction with above speed

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

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, with a standard deviation of σ = 0.5°C. How many readings must be taken so that the probability is 0.90 that the average is within ±0.1◦C of the actual temperature? Round the answer to the next largest whole number.



68 readings.


We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

0.9 = P(|\bar{x}-\mu|<0.1)

0.9 = P(\frac{|\bar{x}-\mu|}{(\sigma)/(√(n))}<(0.1)/((\sigma)/(√(n))))

0.9 = P(|Z|<(0.1)/((\sigma)/(√(n))))

For a confidence level of 90% our Z_(critic) is 1.645


(0.1)/((\sigma)/(√(n))) = 1.645

Substituting for \sigma = 5 and re-arrange for n, we have that n is equal to





We need to make 68 readings for have a probability of 90% and our average is within 0.1\°\frac

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