Why is it harder to hear someone through a wall?

A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the fr

ANEK [815]

The complete question is;

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.

Answer:

A) Amplitude; A = 0.650 m

B) Frequency; f = 1.337 Hz

C) total energy = 17.142 J

D) Kinetic energy = 11.884 J

Potential Energy = 5.258 J

Explanation:

We are given;

Mass;m = 1.15 kg

Equation; x = 0.650 cos (8.40t)

(a) The standard form of a wave function is in the form y(x,t) = Asin(kx−ωt+ϕ)

So, comparing terms in our equation in the question to this, the amplitude is;

A = 0.650 m

(b) we know that formula for frequency is;

f = ω/2π

Again, comparing terms in the standard equation and our question, we can see that ω = 8.4

Thus;

f = 8.4/(2π)

f = 1.337 Hz

(c) Formula for the total energy is given by;

E = m•ω²•A²/2

Plugging in the relevant values, we have;

E = (1.15)(8.40)²(0.650)²/2

E = 17.142 J

(d) we want to find the kinetic energy and potential energy when x = 0.360 m.

The formula for kinetic energy in this case is given by;

K = (1/2)•m•ω²•(A² - x²)

Thus;

K = (1/2) × (1.15) × (8.40)² × ((0.650)² - (0.360)²)

K = 11.884 J

Also, the formula for the potential energy in this case is given by;

U = (1/2)•m•ω²•x²

Thus;

U = (1/2) × (1.15) × (8.40)² × (0.360)²

U = 5.258 J

3 0

3 years ago

Two balls of mass 0.09 kg hang on strings attached to the same point on the ceiling. The balls are given charges Q that cause th

telo118 [61]

Answer:

Q = 6.33μC

Explanation:

To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:

$-Mg+Tcos\theta=0\\\\F_e-Tsin\theta=0$

M: mass of the ball = 0.09kg

T: tension of the string

F_e: electric force between charges

angle = 45°

The electric force is given by:

$F_e=k\frac{Q^2}{r^2}$

Q: charge of the balls

r: distance between balls = 2m

You divide both equation in order to eliminate the tension T:

$tan\theta=\frac{F_e}{Mg}=k\frac{Q^2}{Mgr^2}$

By doing Q the subject of the formula and replacing you obtain:

$Q=\sqrt{\frac{tan\theta Mgr^2}{k}}=\sqrt{\frac{tan45\°(0.09kg)(2m)^2}{(8.89*10^{9}Nm^2/C^2)}}=6.33*10^{-6}C=6.33\mu C$

hence, the charge of the balls is 6.33μC

4 0

3 years ago

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Illusion [34]

<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

v is the velocity in m/s
KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

Based on the problem, the givens are:

KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$

Therefore, the kinetic energy is 9 Joules.

3 0

2 years ago

Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0

3 years ago

Which of the following statements describe the transfer of energy a. Collision of atoms causing nuclear reactions B. A car speed

oksano4ka [1.4K]

I think the answer would be A.

8 0

3 years ago