All categories

3 years ago

I have my exam monday I need some help!

Physics

1 answer:

Lana71 [14]3 years ago

7 0

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

$f=\frac{v}{\lambda}$ and filling in:

$2.5=\frac{v}{2.0}$ so

v = 2.5(2.0) and

v = 5.0 m/s

You might be interested in

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

Why is a normal eye not able to see clearly the object placed closer than 25 CM?

podryga [215]

______________________________

Answer: The nearest position of an object from a normal human eye so that its image is formed on retina is 25 CM. If the object is placed at a distance less than 25 CM, then the blurred image of the object is formed on retina as the focal length of a lens cannot be decreased below a certain limit. Hence we cannot see it clearly.

______________________________

3 0

3 years ago

A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$