Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.
Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,
Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules
Just ignore the negative value for the final result because work is a scalar quantity.
Answer:
Option D
The air pressure inside the car is greater than the pressure outside.
Explanation:
When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.
The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards
Answer:
B
Explanation:
You always want to test as many samples as possible
Answer:
The correct answer is "0.246".
Explanation:
Given that the amplitude is decreased by a factor of 9, then
As we know,
Energy will be:
⇒ 
and,
⇒ 
⇒ 
On putting the estimated values, we get
⇒ 
