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Dima020 [189]
3 years ago
6

Describe a well-known hypothesis that was discarded because it was found to be untrue.earth-centered model of the universe. the

movement of sun, moon, and earth
Physics
1 answer:
polet [3.4K]3 years ago
6 0
The widely accepted hypothesis before that turned out wrong was the Earth-Centered theory or the Geocentric Theory. This was proposed by the philosopher Ptolemy. He came about to this hypothesis from hi observation that from the Earth's perspective, the celestial bodies like the Sun, stars and the moon, look like they rotate around the Earth each day and night. However, this was disproved by Galileo Galelei by his Heliocentric Theory. He observed through the telescope that the Venus also changes phases like the moon. However, he deduced that this is not possible from the positions of the Venus, Earth, Moon and Sun. 
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Richard is driving home to visit his parents. 135{\rm mi} of the trip are on the interstate highway where the speed limit is 65{
nydimaria [60]

Answer:

time spent = 0.2276

Explanation:

given data

distance = 135 mi

usual speed = 65 mph

today speed = 73 mph

solution

we get here time that is express as

time = \frac{distance}{speed}    ...................1

usual time = \frac{135}{65}  = 2.0769 h

today time = \frac{135}{73} = 1.8493 h

so we get here time spent as

time spent = 2.0769 h - 1.8493 h

time spent = 0.2276

6 0
3 years ago
A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe
nataly862011 [7]

Answer:

v_x = 11.78 m/s

Explanation:

Velocity of the ship is given as

v = 12 units

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

v_x = v cos\theta

v_x = 12 cos11

v_x = 11.78 m/s

7 0
3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
3 years ago
PLSSS HELP I WILL GIVE U BRAINLIEST annabelle and jose are building a complete circuit for their science inquiry in class they w
Agata [3.3K]

Answer:

i think its C hope this is rhight have a good day

Explanation:

6 0
3 years ago
Read 2 more answers
A 15 N force and a 45 N force act on in object in opposite directions. What is the net force on the object?
vovikov84 [41]

Answer:

30N in the direction the 45N acts.

Explanation:

Fnet = F1 + F2 (the vector sum of the forces)

Assigning a positive direction to the 45N force and a negative direction to the 15N force gives:

Fnet = 45 - 15

Fnet = 30N

Since the answer is positive, it is in the direction the 45N force acts.

7 0
3 years ago
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