Answer:
the application of scientific knowledge for practical purposes, especially in industry. Another answer:the sum of techniques,skills,methods and processes used in the production of goods or services or in the accomplishment of objectives, such as scientific investigation
Answer: 10.58 C has flowed during the lightning bolt
Explanation:
Given that;
Time of flow t = 1.2 × 10⁻³
perpendicular distance r = 21 m
Magnetic field B = 8.4 x 10⁻⁵ T
Now lets consider the expression for magnetic field;
B = u₀I / 2πr
the current flow is;
I = ( B × 2πr ) / u₀
so we substitute
I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷
= 0.01107792 / 0.000001256
= 8820 A
Hence the charge flows during lightning bolt will be;
q = It
so we substitute
q = 8820 × 1.2 × 10⁻³
q = 10.58 C
therefore 10.58 C has flowed during the lightning bolt
Answer:
mass = 0.18 [kg]
Explanation:
This is a classic problem where we can apply the definition of density which is equal to mass over volume.
![density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]](https://tex.z-dn.net/?f=density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5C%5C%5Cwhere%3A%5C%5Cvolume%20%3D%201%20%5Bm%5E3%5D%5C%5Cdensity%20%3D%200.18%5Bkg%2Fm%5E3%5D)
mass = 0.18*1
mass = 0.18 [kg]
Answer:
T = 1.2 s
T = 15.1 m = 15 m
Explanation:
This is a case of projectile motion:
TOTAL TIME OF FLIGHT:
The formula for total time of flight in projectile motion is:
T = 2 V₀ Sinθ/g
where,
T = Total Time of Flight = ?
V₀ = Launch Speed = 13.9 m/s
θ = Launch Angle = 25°
g = 9.8 m/s²
Therefore,
T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)
<u>T = 1.2 s</u>
<u></u>
RANGE OF BALL:
The formula for range in projectile motion is:
R = V₀² Sin2θ/g
where,
R = Horizontal Distance Covered by ball = ?
Therefore,
T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)
<u>T = 15.1 m = 15 m</u>
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
