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3 years ago

How permeable and porous would an aquifer be?

1 answer:

7 0

Answer: An aquifer is a body of saturated rock through which water can easily move. Aquifers must be both permeable and porous and include such rock types as a sandstone.

Explanation: However, if these rocks are highly fractured, they make good aquifers.

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A circular track has a radius of 32.7 m. What is the centripetal acceleration of a runner moving at the rate of 6.2 m/s?

Jlenok [28]

Given: Radius r = 32.7 m; Velocity V = 6.2 m/s

Required: Centripetal acceleration ac = ?

Formula: ac = V²/r

ac = (6.2 m/s)²/32.7 m

ac = 1.18 m/s²

4 0

4 years ago

A father racing his son has half the kinetic energy of the son, whohas three-fifths the mass of the father. The father speeds up

Afina-wow [57]

Answer:

a) 6.04 m/s

b) 11.02 m/s

Explanation:

a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s

$E_2 = 2E_1$

$\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}$

$V^2 + 5V + 6.25 = 2V^2$

$V^2 - 5V - 6.25 = 0$

$V \approx 6.04m/s$

b) The son kinetic energy initially is:

$E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J$

We can solve for the son speed by the following formula

$E_s = \frac{mv^2}{2}$

$v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s$

$v = \sqrt{121.4} = 11.02 m/s$

5 0

3 years ago

wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o

tresset_1 [31]

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

(b) What is the total charge stored in the parallel combination?

Answer:

i $Q_1 = 2.124 *10^{-11} \ C$

ii $Q_2 = 4.4604 *10^{-11} \ C$

$Q_{eq} = 6.5844 *10^{-11} \ C$

Explanation:

From the question we are told that

The voltage of the battery is $V = 12.0 \ V$

The plate area of each capacitor is $A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2$

The separation between the plates is $d = 2.65 \ mm = 2.65 *10^{-3} \ m$

The permittivity of free space has a value $\epsilon_o = 8.85 *10^{-12} \ F/m$

The dielectric constant of the other material is $z = 2.10$

The capacitance of the first capacitor is mathematically represented as

$C_1 = \frac{\epsilon * A }{d }$

substituting values

$C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 1.77 *10^{-12} \ F$

The charge stored in the first capacitor is

$Q_1 = C_1 * V$

substituting values

$Q_1 = 1.77 *10^{-12} * 12$

$Q_1 = 2.124 *10^{-11} \ C$

The capacitance of the second capacitor is mathematically represented as

$C_2 = \frac{ z * \epsilon * A }{d }$

substituting values

$C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 3.717 *10^{-12} \ F$

The charge stored in the second capacitor is

$Q_2 = C_2 * V$

substituting values

$Q_2 = 3.717*10^{-12} * 12$

$Q_2 = 4.4604 *10^{-11} \ C$

Now the total charge stored in the parallel combination is mathematically represented as

$Q_{eq} = Q_1 + Q_2$

substituting values

$Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}$

$Q_{eq} = 6.5844 *10^{-11} \ C$

7 0

4 years ago

Suppose the balloon is descending with a constant speed of 4.2 m/s when the bag of sand comes loose at a height of 35 m. What is

Margaret [11]

Answer:

2.28 s

Explanation:

Let g = 9.8 m/s2 and neglect air resistance. The box of sand with an initial velocity of 4.2m/s in free fall would yield the following equation of motion

$s = v_0t + gt^2/2$