Asked by PIMP
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
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
Answers
Answered by
PIMP
PLEASE ANSWER. I HAVE PHYSIS REGENTS TOMMORW MUY IMPORTANTE
Answered by
bobpursley
Hmmmm. You have to be a highschool boy. You use the immature name PIMP, and you expect serious responses from adults. Maybe you are just joking about you need help, how are we to know?
Answered by
drwls
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.
(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.
Answered by
personwhowantstolearn
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
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
Answered by
bobpursley
The mass does not change with position.
Answered by
Stace
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.
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.
Answered by
Anonymous
>Uses the name PIMP
>Expects a serious homework question
>Expects a serious homework question
Answered by
anonymous
stace is a real one
Answered by
Jen
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.
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