Answer:
X: period
Y: tangential speed
Explanation:
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(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
(b) The work done in firing the projectile is 2,500 J.
<h3>
Kinetic energy of the projectile at maximum height</h3>
The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;
K.E = ¹/₂mv₀ₓ²
where;
- m is mass of the projectile
- v₀ₓ is the initial horizontal component of the velocity at maximum height
<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.
K.E = (0.5)(2)(30²)
K.E = 900 J
<h3>Work done in firing the projectile</h3>
Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.
W = K.E(i) = ¹/₂mv²
where;
- v is the resultant velocity
v = √(30² + 40²)
v = 50 m/s
W = (0.5)(2)(50²)
W = 2,500 J
Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
The work done in firing the projectile is 2,500 J.
Learn more about kinetic energy here: brainly.com/question/25959744
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1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters
Answer:
979.6 kg/m³
Explanation:
We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²
So, density ρ = P/hg
Since P = 100.8 kPa = 100.8 × 10³ Pa
substituting the values of the variables into the equation for ρ, we have
ρ = P/hg
= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)
= 100.8 × 10³ Pa ÷ 102.9 m²/s²
= 0.9796 × 10³ kg/m³
= 979.6 kg/m³
So, the density of the liquid is 979.6 kg/m³
