Answer:
Correct answer: m = 1,160 kg
Explanation:
Given:
P = 104 hp = 77.48 kW = 77.48 · 10³ W
V₁ = 21 m/s Initial speed (velocity)
V₂ = 29 m/s Final speed (velocity)
t = 3 seconds The time interval during which the work was performed
m = ?
The formula for calculating power is:
P = A / Δt
Since work is a measure of energy change in this case kinetic:
P = ΔEk / Δt = (Ek₂ - Ek₁) / Δt = (m V₂²/ 2 - m V₁²/ 2) / Δt
P = m · (V₂² - V₁²) / 2 · Δt ⇒ m = (2 · P · Δt) / (V₂² - V₁²)
m = (2 · 77.48 · 10³ · 3) / (29² - 21²) = 464.88 · 10³ / (841 - 441)
m = 464.88 · 10³ / 400 = 1.16 · 10³ = 1,160 kg
m = 1,160 kg
God is with you!!!
Answer:
ρ₀ = 3λ/(2πR⁴)
Explanation:
If we take an elemental part of the cross sectional Area of the infinitely long cylinder.
The small elemental part has a small length dr in the cylinder's radial direction such that the cross sectional Area of the elemental part = 2πr dr
The linear density over the entire radial length of the cylinder will be equal to the sum of (volume charge density × elemental cross sectional Area)
That is,
λ = Σ ρ₀rR × 2πr dr
summing from 0 to R
λ = ∫ᴿ₀ (ρ₀rR × 2πr) dr
λ = ∫ᴿ₀ ρ₀R × 2πr² dr
λ = [ρ₀R × 2πr³/3 ]ᴿ₀
λ = (2π ρ₀R/3) [ r³ ]ᴿ₀
λ = (2π ρ₀R/3) [ R³ ]
λ = 2π ρ₀R⁴/3
ρ₀ = 3λ/(2πR⁴)
Answer: Sam provides the biggest power
Explanation: <u>Power</u> is defined, in Physics, as the rate of work done. <u>Work</u> is energy transferred to an object due to the force causing the displacement of the object.
So, to determine Power:
For Sam:
P =
P = 50 Watts
The unit for Power is [P] = J/s = Watt
For Gary:
P =
P = 40 Watts
Comparing power of Sam and Gary, we can conclude that Sam provides the biggest power of 50 Watts.
<h2>
Answer:</h2>
<em><u>(a). v = 7.745 m/s</u></em>
<em><u>(b). a = -1500 m²/s</u></em>
<h2>
Explanation:</h2>
In the question,
(a).
We know that the height of the squirrel from the ground, h = 3 m
Now,
From the equation of the motion we can say that,
Also,
<em><u>Therefore, the velocity of squirrel before hitting the ground is 7.745 m/s.</u></em>
(b).
The squirrel stops in the distance of 2 cm after falling through the height.
So,
The deceleration of the squirrel is = a
So, using the law of the motion in the last 2 cm of distance travelled.
Where,
u = 7.745 m/s
So,
<em><u>Therefore, the deceleration of the squirrel is given by, a = -1500 m²/s</u></em>