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yanalaym [24]
3 years ago
13

if the Horse and Rider have a combined mass of 572 kg what force would be required to accelerate them 5 kph per second

Physics
1 answer:
Vlad [161]3 years ago
8 0

Answer:

   Force required to accelerate = 794.44 N

Explanation:

 Force required = Mass of horse x Acceleration of horse

 Mass of horse and rider, m=   572 kg

 Acceleration of horse and rider, a = 5 kph per second

                                      =\frac{5*1000}{60*60} =1.39 m/s^2

  Force required = ma

                             = 572 x 1.39 = 794.44 N

  Force required to accelerate = 794.44 N

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Also, the amount of activation energy gives an idea of the external energy required to initiate the reaction (for example, by heating the reactants).
Furthermore, by the same principle, we can also deduce the activation energy for the reverse reaction.

If a catalyst is available, the diagram will show a reduced activation energy, compared to a reaction without catalyst.  However, it will also show that the catalyst does not alter the initial and final energies of the reaction.
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How many valence electrons are in an atom of fluorine
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3 years ago
Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis
WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{Mm}{r^{2}} (1)

Where:

F is the gravitational force between Earth and Moon

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M=5.972(10)^{24} kg is the mass of the Earth

m=7.349(10)^{22} kg is the mass of the Moon

r=3.9(10)^{8} m is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to F:

F=m.a_{C} (2)

Where a_{C} is the centripetal acceleration given by:

a_{C}=\frac{V^{2}}{r} (3)  

Being V the orbital velocity of the moon

Making (1)=(2):

m.a_{C}=G\frac{Mm}{r^{2}} (4)

Simplifying:

a_{C}=G\frac{M}{r^{2}} (5)

Making (5)=(3):

\frac{V^{2}}{r}=G\frac{M}{r^{2}} (6)  

Finding V:

V=\sqrt{\frac{GM}{r}} (7)

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}} (8)

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V=1010.92 m/s

5 0
2 years ago
A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t
Mazyrski [523]

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³\rightarrow 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

                                                    \boxed{\boxed{\textbf{d = v * t} } }

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of <u>257,33 kilometers for hour.</u>

<u></u>

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

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¿Good Luck?

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Answer:

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