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Tema [17]
3 years ago
5

The 2-kg spool s fits loosely on the inclined rod forwhich the coefficient of static friction is s= 0.2. if thespool is located

0.25 m from a, determine the minimumand maximum constant speed the spool can have so that it does not slipalongthe rod.

Physics
1 answer:
Aleks [24]3 years ago
7 0
Check the attached file for the answer.

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Hi I’m having some difficulty with this question could really use some help!
Serhud [2]
\begin{gathered} v=\lambda f \\ v=4m\cdot2s^{-1} \\ v=8m/s \end{gathered}

3 0
1 year ago
What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the
alisha [4.7K]

Answer:

1.1648×10⁻¹¹ N

Explanation:

Using

F = qvBsinФ..................... Equation 1

Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.

Note: The angle between v and B = 90°

Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°

Substitute into equation 1

F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°

F = 11.648×10⁻¹²

F = 1.1648×10⁻¹¹ N.

6 0
3 years ago
Read 2 more answers
During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the
natita [175]

Answer:

Explanation:

If the work done on the cart is NET work

Then the work will result in an increase in kinetic energy

KE₀ + W = KE₁

½mv₀² + W = ½mv₁²

½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²

v₁ = 1.626991...

v₁ = 1.6 m/s

4 0
2 years ago
What is the primary evidence used to determine how the Moon formed? O A. Moon craters and Earth craters were caused by the same
Leona [35]
Answer is a
Explanation it is a
7 0
2 years ago
Read 2 more answers
rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon
Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0
3 years ago
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