1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic
1 answer:
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0.04m²
Explanation:
Given parameters:
Pressure = 250000Pa
Weight = 40000N
Unknown:
Area of each foot = ?
Solution:
Pressure is the force exerted per unit area of a body
Pressure =
To find the area;
Area =
Area = = 0.16m²
The force exerted by all the four feet is 0.16m²
the area of each feet = = 0.04m²
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Refer to the diagram shown below.
Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s
The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.
Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)
Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)
Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s
The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.
Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)
Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)