To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then
Third resonance occurs at
At resonance
Therefore at 1m will the tuning fork again create a resonance condition
The ball's horizontal and vertical velocities at time are
but the ball is thrown horizontally, so . Its horizontal and vertical positions at time
are
The ball travels 22 m horizontally from where it was thrown, so
from which we find the time it takes for the ball to land on the ground is
When it lands, and
Answer:
the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s
Explanation:
given information
car's mass, m = 1200 kg
= 100 m
=
= 150 m
= 0
according to conservative energy
the distance from point A to B, h = 150 m - 100 m = 50 m
the initial speed
final speed = 0
thus,
² =
² - 2 g h
0 = ² - 2 g h
² = 2 g h
= √2 g h
= √2 (9.8) (50)
= 31.3 m/s
Answer:
Proton: v=0.689 m/s
Neutron: v=0.688 m/s
Electron: v=1265.078 m/s
Alpha particle: v=0.173 m/s
Explanation:
De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:
λ=
h is the Planck constant: 6.626×10⁻³⁴
We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:
λ=
v=h÷(mλ)
<u>Proton:</u>
m=1.673×10⁻²⁴ g · =1.673×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)
v=0.689 m/s
<u>Neutron:</u>
m=1.675×10⁻²⁴ g · =1.675×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)
v=0.688 m/s
<u>Electron:</u>
m= 9.109×10⁻²⁸ g · =9.109×10⁻³¹ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(9.109×10⁻³¹ kg×575×10⁻⁹m)
v=1265.078 m/s
<u>Alpha particle:</u>
m=6.645×10⁻²⁴ g · =6.645×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)
v=0.173 m/s