The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin
θ.
The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.
If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.
Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
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Answer:
1.3 × 10⁸ e⁻
Explanation:
When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:
20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻
1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:
2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻
Answer:
<h2>1116.9 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 438 × 2.55
We have the final answer as
<h3>1116.9 N</h3>
Hope this helps you
Answer:
a) -41.1 Joule
b) 108.38 Kelvin
Explanation:
Pressure = P = 290 Pa
Initial volume of gas = V₁ = 0.62 m³
Final volume of gas = V₂ = 0.21 m³
Initial temperature of gas = T₁ = 320 K
Heat loss = Q = -160 J
Work done = PΔV
⇒Work done = 290×(0.21-0.62)
⇒Work done = -118.9 J
a) Change in internal energy = Heat - Work
ΔU = -160 -(-118.9)
⇒ΔU = -41.1 J
∴ Change in internal energy is -41.1 J
b) V₁/V₂ = T₁/T₂
⇒T₂ = T₁V₂/V₁
⇒T₂ = 320×0.21/0.62
⇒T₂ = 108.38 K
∴ Final temperature of the gas is 108.38 Kelvin
