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Afina-wow [57]
3 years ago
9

When studying atoms, scientists can ignore the ______ force between charged particles that make up the atoms because it is many

millions of times smaller than other forces in the atom.
Gravitational
Constant
Normal
Fricitional
Physics
2 answers:
MArishka [77]3 years ago
6 0

Answer:

Gravitational

Explanation:

When studying atoms, scientists can ignore the  gravitational force between charged particles that make up the atoms because it is many millions of times smaller than other forces in the atom.

Reptile [31]3 years ago
4 0

Answer:

Gravitational

Explanation:

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3 years ago
An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra
slega [8]

Answer:

Part a)

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

E = 4.4 \times 10^{-13} J

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

m_1v_1 + m_2v_2 + m_3v_3 = 0

(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0

(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

By equation of kinetic energy we have

E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2

E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}

E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}

E = 4.4 \times 10^{-13} J

8 0
3 years ago
Define the Element of Art, value. If you were using a pencil, how would you create a darker value?
ankoles [38]

Elements of art are stylistic features that are included within an art piece to help the artist communicate. The seven most common elements include line, shape, texture, form, space, colour and value, with the additions of mark making, and materiality.

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8 0
3 years ago
The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\
lbvjy [14]
 it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt 
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so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ] 
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt 
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases 
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume 
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ] 
5 0
3 years ago
When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s
d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(2)^2 = 0.5 J

2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(4)^2 = 2 J

4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(5)^2 = 3.125 J

5) when speed is 6 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(6)^2 = 4.5 J

3 0
3 years ago
Read 2 more answers
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