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arsen [322]
3 years ago
5

Why does the frequency of a siren get higher as an ambulance using that siren gets closer?

Physics
2 answers:
Archy [21]3 years ago
7 0
The answer is D. As the ambulance gets closer, the sound waves are compressed relative to the person; so the frequency increases.

yarga [219]3 years ago
7 0
I would say that the answer is D)
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550 g of water at 105°C is poured into an 855 g aluminum container with an initial temperature of 11°C. The specific heat of alu
nexus9112 [7]

Answer:

54.22 kJ

Explanation:

In this case, we need to calculate the heat. The expression to use is:

Q = m * C * ΔT

Now, with the specific heat of water (4186 J/ kg K), we can calculate the temperature in which this occurs.

So:

Q = 0.550 * 4186 * (105 - T)

Q = 2302.3 (105 - T)

Q = 241,741.5 - 2302.3T

Now with the Aluminium:

Q = 0.855 * 900 * (T - 11)

Q = 769.5T - 8464.5

Now, with both equations, we solve for the final temperature:

769.5T - 8464.5 = 241,741.5 - 2302.3T

(2302.3 + 769.5)T = 241,741.5 + 8464.5

3071.8T = 250,206

T = 81.45 K

This is the temperature which the change occurs. Now, let's determine the amount of heat from water to Al:

Q = 241,741.5 - 2302.3(81.45)

Q = 54,219.17 J or simply 54.22 kJ.

5 0
3 years ago
DO I JUST DRAW A SIMPLE WATER MOLECULE? Sketch a space filling model of a water molecule, show polarity by label positive and ne
Sophie [7]
You sketch a water molecule and show a partial positive charge on the two Hydrogen atoms and a partial negative charge on the oxygen atom.
6 0
3 years ago
Time (min)
charle [14.2K]

Answer:

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8 0
3 years ago
Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
zimovet [89]
The question is asking to calculate the object's speed v1, v2, v3 at the certain time is the given of the problem, in my calculation, I would say that the speed would be  2m/s, 1.5m/s, 0.22m/s. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
7 0
3 years ago
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
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