All categories

3 years ago

A wave has a frequency of 46 Hz and a wavelength of 1.7 meters. What is the speed of this wave?

Physics

2 answers:

Sunny_sXe [5.5K]3 years ago

8 0

Answer:

v = 78.2m/s

Explanation:

Check attachment below

Andrej [43]3 years ago

7 0

Explanation:

wavespeed=frequency x wavelength

you only need to 46hz*1.7m which equals

78.2m/s

You might be interested in

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

3 years ago

The majority of early psychological research reflected the __________.

KIM [24]

Differences among males and females

8 0

3 years ago

A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a

DanielleElmas [232]

Answer:

a) $f_o=454.11Hz$

b) $f_o=425.89Hz$

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

$f_o=\frac{v\pm v_o}{v\pm v_s} *f_s$

where:

$f_o=Observed\hspace{3}frequency$

$f_s=Actual\hspace{3}frequency$

$v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves$

$v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source$

$v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer$

Now let's consider the next cases:

$+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source$

$-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source$

$-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer$

$+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer$

The data provided by the problem is:

$f_s=440Hz\\v_o=11m/s$

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

$v=343m/s$

Now, in the first case the observer alone is in motion towards to the source, hence:

$f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz$

Finally, in the second case the observer alone is in motion away from the source, so:

$f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz$

6 0

4 years ago

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil

Oksanka [162]

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

$p_f=p_i$

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

$m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}$ (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

$0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}$

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

6 0

4 years ago

Ice cube is dropped into a glass of water. Describe the motion of the ice cube and explain why it moves this way

krek1111 [17]

The motion of the ice cube: falling. it falls because you move it with force and drop it that makes it fall.

3 0

4 years ago

Read 2 more answers