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Lera25 [3.4K]
4 years ago
11

The atomic mass of an atom (in amu) is equal to

Physics
1 answer:
ladessa [460]4 years ago
5 0

An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus. In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass.

To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons.

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A 1.0 kg ball at the end of a 2.0 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 10
Trava [24]

Answer:

a) 4.65m/s

b) 59.8 N , 1.01125 N

Explanation:

a)

m = mass of the ball = 1 kg

r = length of the string = 2.0 m

h = height gained by the ball as it moves from lowest to topmost position = 2r = 2 x 2 = 4 m

v = speed at the lowest position = 10 m/s

v' = speed at the topmost position = ?

Using conservation of energy

Kinetic energy at topmost position + Potential energy at topmost position = Kinetic energy at lowest position

(0.5) m v'² + m g h = (0.5) m v²

(0.5) v'² + g h = (0.5) v²

(0.5) v'² + (9.8 x 4) = (0.5) (10)²

v' = 4.65m/s

b)

T' = Tension force in the string when the ball is at topmost position

T = Tension force in the string when the ball is at lowest position

At the topmost position:

force equation is given as

mg + T' = \frac{m v'^{2}}{r}

(1)(9.8) + T' = \frac{(1) (4.65)^{2}}{2}

T' = 1.01125 N

At the lowest position:

force equation is given as

T - mg = \frac{m v^{2}}{r}

T - (1) (9.8) = \frac{(1) (10)^{2}}{2}

T = 59.8 N

8 0
3 years ago
The height above the ground of a stone thrown upwards is given by​ s(t), where t is measured in seconds. After 1secondnothing​,t
RoseWind [281]

Answer:

v = 14 ft/s

Explanation:

  • By definition, the average velocity, is just the rate of change of the position, with respect to time, which can be written as follows:

       v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_{f}-x_{o}}{t_{f} - t_{o}}  (1)

  • Defining the vertical position as the y-coordinate, with the origin at ground level, and the upward direction as positive, we can write (1) as follows:

       v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}}  (2)

  • where yf = 54 ft, y₀ = 47 ft, tif = 1.5 s, t₀ = 1s.
  • Replacing in (2) we get:

       v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} = \frac{7ft}{0.5s} = 14 ft/s (3)

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