All categories

2 years ago

A wave has wavelength of 8m and a speed of 360 m/s. What is the frequency of the wave?

Physics

1 answer:

Naily [24]2 years ago

8 0

Answer:

$45~ \text{Hz}.$

Explanation:

$\text{Frequency,}~ f = \dfrac{v}{\lambda}\\\\\\~~~~~~~~~~~~~~~~~~=\dfrac{360~ \text{ms}^{-1}}{8~ \text m}\\\\\\~~~~~~~~~~~~~~~~~~=45~ \text s^{-1}\\\\\\~~~~~~~~~~~~~~~~~~~=45~ \text{Hz}$

You might be interested in

A 1800-kg Jeep travels along a straight 500-m portion of highway (from A to B) at a constant speed of 10 m/s. At B, the Jeep enc

vladimir1956 [14]

Answer:

3600N

Explanation:

$F=m\frac{v^2}{r}$

4 0

3 years ago

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

marshall27 [118]

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1+P_2V_2=P_fV_f$

where,

$P_1$ = first pressure = 8.19 atm

$P_2$ = second pressure = 2.65 atm

$V_1$ = first volume = 2.14 L

$V_2$ = second volume = 9.84 L

$P_f$ = final pressure = ?

$V_f$ = final volume = 2.14 L + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

$8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L$

$P_f=3.64atm$

Therefore, the final pressure of the system in atm is, 3.64 atm

4 0

3 years ago

Draw the following patterns as they would appear when viewed through a compound microscope.:

goblinko [34]

2 j x b
H c e c hope this helppppsssssssss

3 0

3 years ago

Read 2 more answers

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an

Andreas93 [3]

Answer:

170 N

Explanation:

Given in the question that, work a bulldozer can do = 4500 J

We will use trigonometry identity to find the distance bulldozer will travel up the hill

sin(35) = opp/hypo

sin(35) = 15/hypo

hypo = 15/sin(35)

hypo = 26.15m

Formula to use

work done = force × distance

Plug values in the above formula

4500 = force x 26.15

force = 4500/26.15

force = 172.08

force ≈ 170 N

3 0

3 years ago

Read 2 more answers