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mr Goodwill [35]
3 years ago
9

Deducing the acceleration = deduce the gradient of velocity-time graphs true or false?

Physics
1 answer:
gavmur [86]3 years ago
6 0

Answer:

True

Explanation:

Understanding the relationship between change in velocity with time will explain to you how the object is accelerating or decelerating. This means acceleration is a ratio calculated from the change in velocity of an object and change in time. In a velocity time graph, the y-axis represents the velocity and the x-axis represents the time.

The slope of the graph m = Δ y-axis values/Δ x-axis values

m= Δvelocity / Δ time

m= Δ v / Δ t ------this the definition of acceleration so;

a= Δ v / Δ t

Conclusion : The slope/gradient of a velocity -time graph is acceleration.

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Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
3 years ago
Decomposition breaks down organic material into simpler units and:
Lady_Fox [76]

Answer:

d.) provides proteins for plants​

Explanation:

5 0
3 years ago
Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 time
skad [1K]

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

<u>g' = 13.5 m/s²</u>

6 0
3 years ago
The period of a satellite in a circular orbit<br> as the distance from the central object increases.
kow [346]

Answer:

increases

Explanation:

4 0
3 years ago
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How long does it take a message to travel from earth to a spacecraft at mars at its farthest from earth (about 400 million km)?
seraphim [82]
<span>A message needs to be traveled to mars from earth. Distance between Earth and Mars (given) s = 400 million km
 Speed of light = 3.00—10^5 km/ sec
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5 0
4 years ago
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