Explanation:
"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."
(556 x 0.68) = static friction of 378.08N. before movement occurs.
The forces (a) and (b) will not move it.
Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.
Forces (c) and (d) will move it, and accelerate it.
Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.
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Each capacitor carry the same charge 'q'.
Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:
= V₁ +V₂ +V₃
This voltage may also be calculated using capacitance and charge;
V = Q/ C
= V₁ +V₂ +V₃
Provided that the total charge is 'q', hence the total voltage can be expressed as:
= (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)
Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.
Learn more about the capacitor here:
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Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg
