The graph below represents the relationship between speed and time for an object moving along a straight line. What is the total
1 answer:
The distance travelled by the object during the first 4 seconds is 80 m
<h3>Definition of speed </h3>Speed is defined as the distance travelled per unit time. Mathematically, it can be expressed as:
Speed = distance / time
With the above formula, we can obtain the distance travelled by the object in the first 4 seconds.
<h3>How to determine the distance travelled </h3>- Speed = 20 m/s
- Time = 4 s
- Distance =?
Speed = distance / time
20 = distance / 4
Cross multiply
Distance = 20 × 4
Distance = 80 m
Complete question:
See attached photo
Learn more about speed:
You might be interested in
Answer:
Explanation:
In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.
The equation given by Newton tells us that
In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.
In the case of , we understand that they are constant.
We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.
In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.
In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line
Answer:
the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.
Explanation:
Mass of satellite, m
orbit radius of first, r1 = r
orbit radius of second, r2 = 2r
Centripetal force is given by
Where v be the orbital velocity, which is given by
So, the centripetal force is given by
where, g bet the acceleration due to gravity
So, the centripetal force
Gravitational force on the satellite having larger orbit
.... (1)
Gravitational force on the satellite having smaller orbit
.... (2)
Comparing (1) and (2),
F' = 4 F
So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.