A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet.
Compared to the acceleration due to gravity near Earth’s surface, the acceleration due to gravity near the
surface of the planet is approximately
(1) the same (3) one-half as great
(2) twice as great (4) four times as great
bobpursley, instead of wasting time on speculating about the poster, just actually help out. Clearly this person needs help because he deliberately came on the site, asking for help. Immature name or not.
Anyway, the answer is one half as great. Here's how you find out:
First the givens are:
vi=0 (b/c it starts at rest)
m= 5kg
d= 22m
t= 3s
a on Earth always equals 9.81 m/s*s
Given these variable, I surmised that the best formula to use is d=vi(t)+(1/2)at^2.
When you plug in the variables given, you are left with:
22=(1/2)a(3)^(2) ,which gives you:
22=4.5(a)
Now its basic algebra. Divide both sides by 4.5 and you end up getting a=4.8 repeating. When you divide 9.81 by 2, you get 4.905, which is close enough, showing you that it one half as great.
Hope that helps! And good luck on the regents.
stace is a real one
So what's up with those people complaining about the username being PIMP?????? I'm pretty sure those people have a much more serious problem because if you are not gonna help this kid then just leave. It's the user's freedom of right to choose whatever username he/she wants to use.
PLEASE ANSWER. I HAVE PHYSIS REGENTS TOMMORW MUY IMPORTANTE
7 On the surface of Earth, a spacecraft has a mass
of 2.00 × 104 kilograms. What is the mass of the
spacecraft at a distance of one Earth radius
above Earth’s surface?
(1) 5.00 × 103 kg (3) 4.90 × 104 kg
(2) 2.00 × 104 kg (4) 1.96 × 105 kg
To determine the acceleration due to gravity near the surface of the planet compared to Earth's surface, we can use the equation of motion:
s = ut + (1/2) at^2
Where:
s = Distance fallen
u = Initial velocity (which is 0, since it starts from rest)
t = Time taken to fall
a = Acceleration due to gravity
In this case, the distance fallen (s) is 22 meters and the time taken to fall (t) is 3.0 seconds. Plugging in these values into the equation, we get:
22 = 0 + (1/2) a(3.0)^2
Simplifying the equation, we have:
22 = (1/2) a(9.0)
Now, let's solve for 'a'. Multiply both sides of the equation by 2/9:
(22 * 2/9) = a
a = 4.89 m/s^2
Now, we have the acceleration due to gravity near the surface of the planet, which is approximately 4.89 m/s^2.
To compare it with the acceleration due to gravity near Earth's surface, we need to know the standard value of gravity on Earth, which is approximately 9.8 m/s^2.
Comparing the two values, we can see that the acceleration due to gravity near the surface of the planet is about one-half as great as the acceleration due to gravity near Earth's surface.
Therefore, the answer is (3) one-half as great.
The mass does not change with position.
On earth, the distance an object would fall in 3.0 seconds is
(g/2)*t^2 = 4.9*9 = 44.1 m
On planet X, it falls 22 meters.
The distance it falls in a given time t is proportional to the acceleration of gravity.
Reach your own conclusion from that.
>Uses the name PIMP
>Expects a serious homework question