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stiv31 [10]
3 years ago
10

What is the mass of an object moving at a speed of 9.0 m/s with a kinetic energy of 200 J?

Physics
1 answer:
Yuliya22 [10]3 years ago
8 0
Ek= 1/2mv²
200=1/2 m(9 ²)
200=1/2(81)m
200=40.5m
m=200/40.5
m=4.93kg

Answer: 4.9 kg (second option)
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Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway
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Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s

So, the velocity of the cart is 12.12 m/s.

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a 64 l gas tank is filled completely is 15 degree celsius how much gasoline overflows into can get up to 35 degree celsius while
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Answer:

Volume of gasoline overflow(v)= 40/9 L (I.e. 4.44 L)

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.....overflow(V)=v2-v1

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3 0
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What is the net charge of a copper atom if it gains 2 electrons?
Alex_Xolod [135]
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7 0
3 years ago
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4 0
3 years ago
Read 2 more answers
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
In-s [12.5K]

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0
3 years ago
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