Answer:
The speed of the bat is 5.02 m/s.
Explanation:
Given that,
Frequency = 30.0 kHz
Frequency of echo = 900 Hz
We need to calculate the frequency
Using formula of beat frequency
Put the value into the formula
We need to calculate the speed of the bat
Using Doppler equation
Put the value into the formula
Hence, The speed of the bat is 5.02 m/s.
The distance between a trough and a crest is double the amplitude.
The distance between trough and crest is (9m - 6m) = 3m.
So the amplitude of those water waves is (3/2) = 1.5 meters.
For this question, we really don't need to know how far apart the crests are, or how often they pass. There may be more parts to the question, for which that information is needed.
For example:
-- What's the wavelength of the waves ? 8 meters.
-- What's the period of the waves ? 4 seconds.
-- What's the frequency of the waves ? 1 / (4 seconds) = 0.25 Hz.
-- What's the speed of the waves ? (8 m) x (0.25 Hz) = 2 m/s
Answer:
r=2.6 cm
Explanation:
Hi!
Lets call steel material 1, and the new alloy material 2. You know their densities:
The volume of a sphere with radius r is given by:
Then the masses of the bearings are:
For the masses to be the same:
![\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm](https://tex.z-dn.net/?f=%5Crho_1%20V_1%20%3D%20%5Crho_2%20V_2%5C%5C%5Cfrac%7BV_2%7D%7BV_1%7D%20%3D%5Cfrac%7B%5Crho_1%7D%7B%5Crho_2%7D%5C%5C%28%5Cfrac%7Br_2%7D%7Br_1%7D%29%5E3%20%3D%20%5Cfrac%7B8.08%7D%7B3.14%7D%3D2.57%5C%5Cr_2%3D%5Csqrt%5B3%5D%7B2.57%7D%5C%3Br1%20%3D1.37r_1%3D1.37%2A1.9cm%3D2.6cm)
Answer:
Explanation:
range of projectile = u² sin2θ / g
u = 30 m /s
θ = 40°
range = 30² x sin 80 / 9.8
= 90.44 m
b )
maximum altitude H = u² sin²θ / 2 x g
= 30² sin²40 / 9.8
= 37.94 m .
