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3 years ago

Apollo astronauts took a "nine iron" to the Moon and hit a golf ball about 180 m.

Physics

2 answers:

Studentka2010 [4]3 years ago

7 0

So in calculating this one its is really hard to explain how i get it on solve it but you must consider this factors that i give in getting the answer. First is the distance cover by the ball when it is hit by the club, Second is you must estimate both of those data when it is in the moon and in the earth whre the gravity of the earth is 9.8m/s^2 so by calculating the Gravity of the moon or gMoon is equal 1.74m/s^2

Andrews [41]3 years ago

4 0

The acceleration due to gravity on the surface of the Moon is 1.7 m/s²

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Let us now tackle the problem!

<u>Given:</u>

$d_{moon} = 180 ~ m$

$d_{earth} = 32 ~ m$

$g_{earth} = 9.8 ~ m/s^2$

<u>Unknown:</u>

$g_{moon} =$ ?

<u>Solution:</u>

The motion of the golf ball is a parabolic motion, then we can use the following formula to calculate the horizontal displacement of the ball.

$\large {\boxed {d = \frac{u^2 \sin 2 \theta}{g}} }$

Let's use this formula to find a comparison of the motion of golf ball on earth and on the moon.

$d_{moon} : d_{earth} = \frac{u^2 \sin 2 \theta}{g_{moon}}} : \frac{u^2 \sin 2 \theta}{g_{earth}}}$

$d_{moon} : d_{earth} = \frac{1}{g_{moon}}} : \frac{1}{g_{earth}}}$

$d_{moon} : d_{earth} = g_{earth} : g_{moon}$

$180 : 32 = 9.8 : g_{moon}$

$g_{moon} = (32 \times 9.8) \div 180$

$\large {\boxed {g_{moon} = 1.7 ~ m/s^2} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Sperm , Whale , Travel

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3 years ago

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

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Answer:

a) t = 746 s

b) t = 666 s

Explanation:

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

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Answer:

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