Answer:
A) About
newtons
B) 76.518 newtons
C) 111.834 newtons
Explanation:
A)
, where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:
![F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}](https://tex.z-dn.net/?f=F_g%3D%5Cdfrac%7B%286.67408%20%5Ctimes%2010%5E%7B-11%7D%29%287.8%29%2811.4%29%7D%7B%280.8%29%5E2%7D%5Capprox%209.273%20%5Ctimes%2010%5E%7B-9%7D)
B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.
C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.
Hope this helps!
Answer:
why you doin this
Explanation:
is this so we get free points?
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:
diesel engine
Explanation:
because diesel is stronger than petrol
Answer:
Gc(s) = ![\frac{0.1s + 0.28727}{s}](https://tex.z-dn.net/?f=%5Cfrac%7B0.1s%20%2B%200.28727%7D%7Bs%7D)
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab