Answer:
v₂ = 176.24 m/s
Explanation:
given,
angle of projectile = 45°
speed = v₁ = 150 m/s
for second trail
speed = v₂ = ?
angle of projectile = 37°
maximum height attained formula,
now,
now, equating both the equations
v₂² = 31061.79
v₂ = 176.24 m/s
velocity of projectile would be equal to v₂ = 176.24 m/s
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
Answer:
Writing reactants on the left, and the products are written on the right
Explanation:
Answer:
∴ [T]=[WF−1V−1]
Hope this answer is right!!
