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Salsk061 [2.6K]
3 years ago
12

High-frequency sound waves are to ______ as low-frequency sound waves are to ______.

Physics
1 answer:
Elodia [21]3 years ago
8 0
High freq sound waves are mainly used for FM radio stations because they carry more information but cant travel as far, low freq sound waves are used for AM radio stations because they have a lower quality sound but can travel much further
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A horse pulls a plow with a 242 N force for
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Answer: The answer is D

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3 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
4 years ago
A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave
damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

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