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stepan [7]
2 years ago
5

If a train travel from Addis Ababa to Dire Dawa at a constant velocity of 400Km/hr and the to Djibouti at a constant velocity of

V. if the average velocity of 520Km/hr then what is the value of V?----------------
Physics
1 answer:
madreJ [45]2 years ago
4 0

The value of V is 640 km/hr.

<h3>What is Average velocity?</h3>

Average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.

To calculate the value of V from the question, we use the formula below.

  • V' = (V+U)/2............ Equation 1

Making V the subject of the equation

  • V = 2V'-U.............. Equation 2

From the question,

⇒ Given:

  • V' = 520 km/hr
  • U = 400 km/hr

Substitute these values into equation 2

  • V = (520×2)-400
  • V = 640 km/hr

Hence, the value of V is 640 km/hr.

Learn more about average velocity here: brainly.com/question/4931057

#SPJ1

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What are the two main types of energy?
djyliett [7]

Answer:

Fourth option

Explanation:

They're many different types of energy, from chemical and mechanical to heat and solar energy. But the two most basic types of energy are "kinetic and potential energy" or the fourth option. Kinetic energy is the energy an object has when it is in motion, while potential energy is the energy an object has when it's as rest. These two specific types of energies are the most basic and you can even convert them into many different types of energies, like heat or electrical energy.

Hope this helps.

7 0
2 years ago
How much power does it take to do 500 J of work in 10 seconds?
Dmitry_Shevchenko [17]
Power = work/time
  
          = 500/10
 
          = 50J/s or 50 watt 


7 0
3 years ago
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t
Natasha_Volkova [10]

Answer:

v=12.5 i + 12.5 j m/s

Explanation:

Given that

m₁=m₂ = m

m₃ = 2 m

Given that speed of the two pieces

u₁=- 25 j m/s

u₂ =- 25 i m/s

Lets take the speed of the third mass = v m/s

From linear momentum conservation

Pi= Pf

0 = m₁u₁+m₂u₂ + m₃ v

0 = -25 j m  - 25 i m + 2 m v

2 v=25 j   + 25 i m/s

v=12.5 i + 12.5 j m/s

Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s

4 0
3 years ago
Physical Science Lesson 4 unit 2 questions, I need the answers
neonofarm [45]
I know I'm a bit late but just in case you still need some of the answers.

1: Oxygen
2: Change in shape
3: Water breaking down into hydrogen and oxygen
4: I think it's Milk but I might be wrong.

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7 0
3 years ago
Read 2 more answers
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
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