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Pie
4 years ago
14

A book is sitting on a shelf that is 3.0 meters off the ground. Kinetic energy is found using the formula 1/2 times mass times s

peed. Potential energy is found by taking mass times g times height, with g as the constant, 9.8 m/s squared. What is needed to determine the mechanical energy used to move this book?
Physics
1 answer:
Sophie [7]4 years ago
5 0

The only thing in either one of those formulas that we don't know is the mass of the book.

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When the forward and reverse paths of a change occur at the same rate, "<span>B. the system is in equilibrium". This is because there is no net flow of reactants at this time. This stage can be reached quickly or after long periods of time.</span>
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3 years ago
Which of the following statements are true?
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A. False, because the SI unit for frequency is Hertz, 1 hertz equals 1 wave passing a fixed point in 1 second,

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8 0
3 years ago
Two objects attract each other gravitationally with a force of 2.5 x 10^-10N when they are 0.25 m apart. Their total mass is 4.0
In-s [12.5K]

Answer:

M = 3.9406 kg and m = 0.0594 kg

Explanation:

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically it is expressed as follows:

Fg = (G×M×m)/r²    Formula (1)

Where:

Fg is the gravitational force (N)

G is the universal gravitation constant, G = 6.67 × 10⁻¹¹ (N×m²)/kg²

M and m are the masses of the bodies that interact  (kg).

r is the distance that separates them (m).

Known Data

Fg = 2.5 × 10⁻¹⁰ N

r = 0.25 m

G = 6.67 × 10⁻¹¹ (N×m²)/kg²

Problem development

We propose 2 equations

M + m = 4kg

M = 4 - m equation (1)

We replace in formula (1)

2.5 × 10⁻¹⁰ = (6.67 × 10⁻¹¹ × M × m)/(0.25)²

2.5 × 10⁻¹⁰ × (0.25)² = (6.67 × 10⁻¹¹ × M × m)

(2.5 × 10⁻¹⁰ × (0.25)²)/(6.67 × 10⁻¹¹) = M × m

M × m = 0.234 equation (2)

We replace M = 4 - m in equation (2)

(4 - m) × m = 0.234

4m - m² = 0.234

m² - 4m + 0.234 = 0 (quadratic equation)

We apply the formula for the quadratic equation and obtain 2 values for m that meet the conditions:

m = 3.9406 kg or m =  0.0594 kg

We replace m in equation (1)

M = 4 - 3.9406 = 0.0594 kg or M = 4 - 0.0594 = 3.9406

To meet the condition that M + m must give 4 kg, one mass must be equal 3.9406 and the other must equal 0.0594, then:

M = 3.9406 kg and m = 0.0594 kg

6 0
4 years ago
you apply 100.0 N of force to a machine. The machine applies 20.0 N of the force to the load. what is the mechanical advantage o
morpeh [17]
Mechanical advantage is 5
100/20
5 0
4 years ago
Read 2 more answers
A parallel-plate capacitor has plates of area 0.30 m2 and a separation of 2.10 cm. A battery charges the plates to a potential d
Dima020 [189]

Answer:

a) 1.26e^-10F

b) 1.47e^-10F

c) 2.39e^-8C   2.89e^-8C

d) E=4500.94N/C

e) E'=5254.23N/C

f) 100.68V

g) 1.65e^-10J

Explanation:

To compute the capacitance we can use the formula:

C=\frac{k\epsilon_o A}{d}

where k is the dielectric constant of the material between the plates. d is the distance between plates and A is the area.

(a) Before the material with dielectric constant is inserted we have that k(air)=1. Hence, we have:

k=1\\A=0.30m^2\\d=0.021m\\e_o=8.85*10^{-12}C^2/(Nm^2)\\\\C=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{0.021m}=1.26*10^{-10}F

(b) With the slab we have that k=4.8 and the thickness is 4mm=4*10^{-3}m. In this case due to the thickness of the slab is not the same as d, we have to consider the equivalent capacitance of the series of capacitances:

C=(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_1})\\\\\\C_1=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{8.5*10^{-3}m}=3.1*10^{-10}F\\\\C_2=\frac{(4.8)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{4*10^{-3}m}=3.186*10^{-9}F\\\\C=1.47*10^{-10}F

(c)

The charge between the plates for both cases, with the slab is given by:

Q : without the slab

Q': with the slab

Q=CV=(1.26*10^{-10}F)(190V)=2.39*10^{-8}C\\\\Q'=C'V=(1.47*10^{-10F})(190V)=2.79*10^{-8}C\\

(d) The electric field between the plate is given by:

E=\frac{Q}{2\epsilon_o A}

E: without the slab

E': with the slab

E=\frac{2.39*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=4500.94N/C\\\\E'=\frac{2.79*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=5254.23N/C\\

(f) We can assume the system as composed by V=V1+V'+V1 as in (c). By using the equation V=Ed we obtain:

V=2(4500.94)(8.85*10^{-3}m)+(5254.23)(4*10^{-3}m)=100.68V

(g) External work is the difference between the energies of the capacitor before and after the slab is placed between the parallels:

\Delta E=\frac{1}{2}[(1.26*10^{-10}F)(120V)-(1.47*10^{-10})(100.6V)]=1.65*10^{-10}J

5 0
4 years ago
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