Answer:
Magnetic force, 
Explanation:
It is given that,
Velocity of proton, 
Angle between velocity and the magnetic field, θ = 53°
Magnetic field, B = 0.49 T
The mass of proton, 
The charge on proton, 
The magnitude of magnetic force is given by :
So, the magnitude of the magnetic force on the proton is 
. Hence, this is the required solution.
Answer:
0.14 J
Explanation:
The maximum velocity is the amplitude times the angular frequency.
vmax = Aω
ω = vmax / A
ω = (3.2 m/s) / (0.06 m)
ω = 53.3 rad/s
For a spring-mass system:
ω = √(k / m)
ω² = k / m
k = ω²m
k = (53.3 rad/s)² (0.050 kg)
k = 142 N/m
The elastic potential energy is:
EE = ½ kx²
EE = ½ (142 N/m) (0.044 m)²
EE = 0.14 J
Answer:
5 feet
Explanation:
First lets imagine that the values are in meters and kg, to make it easier for me.
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The force created by the teacher is 120 × 5 = 600N.
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The force created by the 50 kg student is 50 × 8 = 400N
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The force that needs to be created by the 40 kg student is 600N - 400N = 200N.
To create that force that student needs to sit 200 ÷ 40 = 5m.
So the 40 lbs student is sitting 5 feet from the fulcrum.
Answer:
- From <u>2,150 kilowatt hours</u> will qualify for the heavy user classification.
Explanation:
Since the monthly electric consumption for January follows a normal distribution, you must use a standard normal table (Z table), which provides the probabilities for the (normalized) Z-scores.
The top 6% of electric consumption, means that you must find the z-score for which the probablitiy is ≥ 0.06.
The corresponding z-score shown in the table is 1.555.
Now just use the equation:
Where:
- μ = mean = 1,650 kilowatt hours
- σ = standard deviation = 320 kilowatt hours
- x = the value of your variable, which you will determine
Substituting:
- 1.555 = (x - 1,650) / 320
- x = 320 × 1.555 + 1,650 = 2,148 ≈ 2,150
- x ≈ 2,150 kilowatt hours.
