Given that,
Frequency emitted by the bat, f = 47.6 kHz
The speed off sound in air, v = 413 m/s
We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :
or
So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.
Answer:
carbon has four unpaired electrons in its valence shell . hydrogen having one unpaired electron in its valence shell comes to bond with carbon by sharing a pair of electrons .since carbon needs 4 electrons to be stable, 4 hydrogen atoms take part in the bond . It's a covalent bond because the difference between the electronegativity of carbon and hydrogen is quite small .
Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
Hello, I see you are in a jam. Lemme help.
1.) True
2.) True
3.) True
4.) True
5.) True
LOL these are all true ;)
Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s
