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telo118 [61]
3 years ago
14

The tired of a car support the weight of a stationary. If onetire has a slow leak, the air pressure within the tire will_____wit

h time, the surface area between the tire and the roadwill____in time, and the net force the tire exerts on the roadwill_____in time.a) Increase, Increase, Increaseb) Decrease, decrease, decreasec) Decrease. increase,increased) Decrease, increase, remain constante) Decrease, increase, decrease
Physics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

d) Decrease, increase, remain constant

Explanation:

If one tire has a slow leak, the air pressure within the tire will_DECREASE____with time due to outflow of air , the surface area between the tire and the road will__INCREASE__in time,due to flattening of tire.

The net force the tire exerts on the road will_REMAIN CONSTANT____in time. It is so because force does not depend upon area. It is pressure which depends upon area. As there is no change in the weight of the car , force on the road will remain constant.

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Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

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Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

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Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

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Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

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