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ella [17]
3 years ago
8

Two balls of equal size are dropped from the same height from the roof of a building. the mass of ball a is twice that of ball b

, m subscript a equals 2 m subscript
b. when the two balls reach the ground, how do their kinetic energies compare?
Physics
2 answers:
Sindrei [870]3 years ago
8 0
The mass of ball a is twice the mass of ball b:
m_a = 2 m_b
This means that the initial potential energy of ball a (U_a = m_a gh=2 m_b gh) is twice the potential energy of ball b (U_b = m_b  g h):
U_a = 2 U_b
When the two balls reach the ground, the potential energy of each ball has converted into kinetic energy (since now their altitude is h=0), because the total mechanical energy of each ball must be  conserved. Therefore:
K_a = U_a
K_b = U_b
and so the kinetic energy of ball a must be twice the kinetic energy of ball b:
K_a = 2 K_b
statuscvo [17]3 years ago
6 0

Answer:

5.5 m/s

Explanation:

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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
What is the density of an object that has a mass of 30 g and a volume of 20cm cubed/ to the third power?
Ksenya-84 [330]

Answer:

d= 1.5 g/cm3

Explanation:

datos

m= 30g

v= 20cm3

d=?

formula

d= m / v

solución

d= 30g / 20cm3 = 1.5g/cm3

6 0
3 years ago
Determine tge energy in joules of a photon whose frequency is 3.55x10
JulijaS [17]
E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J

Hope it helped!
6 0
3 years ago
Select the correct answer.
mamaluj [8]

Answer:

amount of charge on the source charge

6 0
3 years ago
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the
Julli [10]

Answer:

<h2>a) Q = 0.759µC</h2><h2>b) E = 39.5µJ</h2>

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0
3 years ago
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