the answer to this question is A because If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. ... In air, a feather and a ball do not fall at the same rate.
The respiratory system is made up of multiple organs, saying that the lungs is the only required organ is very wrong. While the lungs are probably the most important part of it, the organs that transfer the air are definitely necessary for everything to work.
The main organ that is needed for air to flow from the mouth to the lungs is the trachea, or the windpipe. This connects the back of the mouth to the lungs, without this, no air could get into the lungs, thus rendering them useless.
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth
Explanation:
The period of a simple pendulum is given by the equation
where
L is the length of the pendulum
g is the acceleration of gravity
In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is
where
is the acceleration of gravity on Earth
The period of the pendulum on the Moon is
where
is the acceleration of gravity on the Moon
Calculating the ratio of the period on the Moon to the period on the Earth, we find
Therefore, for every hour interval on Earth, the Moon clock will display a time of
A) (9.8/1.6)h
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