Answer:
7.89 7.91
Explanation:
The ranges of measurement lie between 7.92-0.05 and 7.92+0.05
7.87g and 7.97g
Answer:
3.7 A
Explanation:
Parameters given:
Magnetic field strength, B = 5 * 10^(-5) T
Distance of magnetic field from wire, r = 1.5 cm = 0.015 m
The magnetic field, B, due to a current, I, flowing a wire is given as:
B = (μ₀*I) / 2πr
Where μ₀ = permeability of free space
To get the current, I, we make I the subject of the formula:
I = (2πr * B) / μ₀
I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))
I = 3.7 A
<span>f(x) = 5.05*sin(x*pi/12) + 5.15
First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.
So our function at this point looks like
f(x) = 5.05*sin(x*pi/12)
But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
f(x) = 5.05*sin(x*pi/12) + 5.15
The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
f(x) = 5.05*sin(x*pi/12 + C) + 5.15
where
C = Phase correction offset.
But we don't need it for this problem, so the answer is:
f(x) = 5.05*sin(x*pi/12) + 5.15
Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
Answer:
frictional force = 0.52 N
Explanation:
diameter of turn table (D1) = 30 cm = 0.3 m
mass of turn table (M1) = 1.2 kg
diameter of shaft (D2) = 1.2 cm = 0.012 m
mass of shaft (M2) = 450 g = 0.45 kg
time (t) = 15 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
radius of turn table (R1) = 0.3 / 2 = 0.15 m
radius of shaft (R2) = 0.012 / 2 = 0.006 m
total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft
I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}
I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}
I = 0.0135 + 0.0000081 = 0.0135081
ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s
α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}
torque = I x α
torque = 0.0135081 x (-0.23) = - 0.00311 N.m
torque = frictional force x R2
- 0.00311 = frictional force x 0.006
frictional force = 0.52 N
Answer:
PE= m * g *h
work:
PE= 65kg * 9.8 kg *8,800 m
PE=5605600 m/kg
idk the actual units i forgot
