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2 years ago

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An Astronaut lands on an Earthlike planet and drops a small lead ball with a mass from the top of her spaceship. The point of re

lease is 25 m above the surface of the planet and the ball takes 5 s to reach the ground. The astronaut's mass is 80 kg. Astronaut weight on the planet is
PLEASE NO LINKS, REAL ANSWERS DON'T WASTE TIME.

1 answer:

bearhunter [10]2 years ago

5 0

First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,

s = ut +(gt^2)/2

Distance s = 25m

Time t = 5 s

Velocity u = 0

By putting these values,

25 = 1/2.g.(5)²

g = 2

So the gravity on that planet is 2. Lets find out the weight of the astronaut.

Mass of the astronaut on earth m = 80 kg

Weight of astronaut on earth W = mg = (80)(9.8) = 784 N

Weight of astronaut on earth like planet = (80)(2) = 160 N

x = 160N

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pochemuha

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3 years ago

A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A

Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

$IMA = \frac{d_i}{d_o}$

where

$d_i$ is the input distance

$d_o$ is the output distance

For the pulley system in this problem, $d_i = 3.90 m$ and $d_o = 0.975 m$ , so the IMA is

$IMA=\frac{3.90 m}{0.975 m}=4$

B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

$MA=\frac{F_o}{F_i}$

where $F_o$ is the output force and $F_i$ is the input force. For the pulley system in this problem, $F_i = 375 N$ and $F_o = 1345 N$ , so the MA is

$MA=\frac{1345 N}{375 N}=3.59$

C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

$\eta = \frac{MA}{AMA} \cdot 100$

Therefore, in this case,

$\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%$

3 0

3 years ago

II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st

AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

$a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}$

Since we start counting at 0s and the ball departs from rest, this is just $a=\frac{v}{t}$

So we can write:

$F=ma=\frac{mv}{t}=\frac{gmv}{gt}$

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

$F=\frac{Wv}{gt}=\frac{v}{gt}W$

Substituting our values:

$F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N$

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0

2 years ago

How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.

goldfiish [28.3K]

143m/s if you just perhaps by what you know you'll figure it out

4 0

2 years ago

A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g

Sliva [168]

To solve for force, you need to get the product of mass and acceleration.
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

$a = \frac{vf-vi}{t}$

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.

So we can put it into our formula now:

$a = \frac{0m/s-50m/s}{20ms}$

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.

$20ms$ x $\frac{1s}{1000ms}$ = $\frac{20s}{1000} = 0.02s$

So your new time is 0.02s. Now we put this time into the formula:

$a = \frac{0m/s-50m/s}{0.02s}$
$a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2}$

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.

So now we have our acceleration. Now using this, we can get our force.

$F= ma$

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is $kg.m/ s^{2}$ but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.

$150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg$

Our new mass is 0.150kg.

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 $m/ s^{2}$

Now that all our units match, we can put that into our formula:

$F= ma$
$F= (0.150kg)(-2,500m/s^{2})$
$F = -375kg.m/ s^{2} or -375N$

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.

4 0

3 years ago