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Juliette [100K]

2 years ago

14

Priscilla is driving her car on a busy street and Harvey passes her on his motorcycle. What will happen to the sound from his mo

torcycle after it
passes her car?

1 answer:

Marrrta [24]2 years ago

6 0

Answer:

what will happen to the sound is the wavelength increases thus frequency decreases so the sound decreases as he move away from her car..

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What are the two major characteristics that change when air is heated and cooled ? How do these characteristics change the movem

sleet_krkn [62]

Two major characteristics that change when air is heated and cooled is the density and the direction of the air flow. They change the movement of the air by heated air rises and cooled air doesn't.

8 0

2 years ago

Newer aircraft jet catapult systems use magnets instead of steam. The launch still takes 1.19 seconds, but the acceleration is a

Westkost [7]

Answer:

33.6 m

Explanation:

Given:

v₀ = 0 m/s

a = 47.41 m/s²

t = 1.19 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (1.19 s) + ½ (47.41 m/s²) (1.19 s)²

Δx = 33.6 m

8 0

2 years ago

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

3 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

$y=14.2* (0.9172535661)^{(31.8)}$ $y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}$

$y=14.2* 0.0641450581$ $y=14.2*(0.5)^{3.962518068}$

$y=0.9108598257$ $y=14.2* 0.0641450581$

$y=0.9108598257$

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0

1 year ago