Answer:
1) λ = 0.413 m
, 2)v = 25,213 m / s
, 3) T = 0.216 N
, 4) m = 22.04 10-3 kg
Explanation:
1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related
λ = 2L / n n = 1, 2, 3 ...
In this case L = 0.62 m and n = 3
Let's calculate
λ = 2 0.62 / 3
λ = 0.413 m
2) the velocity related to wavelength and frequency
v = λ f
v = 0.413 61
v = 25,213 m / s
3) let's use the equation
v = √T /μ
T = v² μ
T = 25,213² 3.4 10⁻⁴
T = 0.216 N
4) the rope tension is proportional to the hanging weight
T-W = 0
T = W
W = m g
m = W / g
m = 0.216 / 9.8
m = 22.04 10-3 kg
5) n = 2
λ = 2 0.62 / 2
λ = 0.62 m
6) v = λ f
v = 0.62 61
v = 37.82 m / s
7) T = v² μ
T = 37.82² 3.4 10⁻⁴
T = 0.486 N
8) m = W / g
m = 0.486 / 9.8
m = 49.62 10⁻³ kg
9) n = 1
λ = 2 0.62
λ = 1.24 m
v = 1.24 61
v = 75.64 m / s
T = v² miu
T = 75.64² 3.4 10⁻⁴
T = 2.572 10⁻² N
m = 2.572 10⁻² / 9.8
m = 262.4 10⁻³ kg
Answer:
Driver hits the dog
Explanation:
From kinematics equation
and making a the subject
F=ma=800*-4=-3200 N
The driver is required to apply a force of 3200 N to effectively not hit the dog. However, when the driver applies 2000 N, this is less the required force by 1200 N hence the driver hits the dog.
Answer:
L = m v r (The momentum remains constant)
Explanation:
Even in an ellipsoidal orbit, the law of conservation of angular momentum always apply. When the plant approached the perihelion, the radius of the orbit decreases and the speed of the star increases to conserve the momentum. Similarly, when the planet approaches the aphelion, the speed of the star decreases as the radius increases to conserve the momentum. So, the momentum at a particular instant can be calculated by L = m v r
The frequency of light wave is 0.6*10^15 Hz.
The relation between the speed of light, frequency, and wavelength is given as
c=λν
Plugging the values in the above equation
3*10^8=5*10^(-7)*ν
ν=0.6*10^15 Hz.
Therefore the frequency of the light wave is 0.6*10^15 Hz.
