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Ronch [10]
3 years ago
10

What equation can be used to solve for acceleration

Physics
1 answer:
Tanya [424]3 years ago
3 0

Answer: acceleration can be calculated when a known force is acting on an object of known mass. Newton’s law can be represented by the equation F net = m x a, where F net is the total force acting on the object, m is the object’s mass, and a is the acceleration of the object.

Explanation:

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Describe what happens to the energy of a wave if the frequency decreases and the frequency increased
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"As frequency increases, wavelength decreases. Frequency and wavelength are inversely proportional. This basically means that when the wavelength is increased, the frequency decreases and vice versa. Wavelength is described as the distance between a trough to a trough or a crest to a crest."

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To run a centrifuge correctly, you should:
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Balance tubes by spacing them equally around the centrifuge and Always balance tubes with other tubes containing a same volume of liquid are right. 
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Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r  were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U =  -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

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7 0
3 years ago
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  • \underbrace{\sf{Elementary~ charge} }

Explanation:

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6 0
3 years ago
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The atomic radii of Li+and O2−ions are 0.068 and 0.140 nm, respectively.(a) Calculate the force of attraction between these two
sasho [114]

Answer:

Part a)

F_{attraction} = 1.06 \times 10^{-8} N

Part b)

F_{repulsion} = 1.06 \times 10^{-8} N

Explanation:

Part a)

Equilibrium distance is the distance between two centers when two ions just touch each other

So here we will have

r = 0.068 nm + 0.140 nm

r = 0.208 nm

Now Force of attraction between two ions is given by Coulomb's law

F = \frac{kq_1q_2}{r^2}

F = \frac{(9\times 10^9)(1.6 \times 10^{-19})(2\times 1.6 \times 10^{-19})}{(0.208\times 10^{-9})^2}

F_{attraction} = 1.06 \times 10^{-8} N

Part b)

At equilibrium separation net force between two ions must be zero

F_{attraction} = F_{repulsion}

so the attraction force and repulsion force must be of equal magnitude

so we have

F_{repulsion} = 1.06 \times 10^{-8} N

6 0
3 years ago
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