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borishaifa [10]

4 years ago

7

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re

sts on a nearly frictionless plane, which is tilted at an angle of 40.0∘ as shown.

1 answer:

slamgirl [31]4 years ago

3 0

Answer:

tension: 19.3 N
acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

mass A = 2.0 kg

mass B = 3.0 kg

θ = 40°

<u>Find</u>

The tension in the string

The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

F/m = a = F/m

(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

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kati45 [8]

Answer:

Explanation:

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Force of pull, F = 31 N

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W = F x d x Cos θ

W = 31 x 53 x cos 40

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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Brrunno [24]

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Find A and effective resistance.<br> (Ill give Brainliest if you provide explaination)

Alex777 [14]

Answer:

A = 2.4 A

$R_{eq} = 5 \ \Omega$

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

$R_{series \, 1}$ = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

$R_{series \, 2}$ = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}$

Therefore;

$R_{eq} = \dfrac{R_{series \, 1} \times R_{series \, 2}}{R_{series \, 1} + R_{series \, 2}}$

Therefore;

$R_{eq} = \dfrac{10\times 10}{10 + 10} \ \Omega = 5 \ \Omega$

$R_{eq} = 5 \ \Omega$

The value of the current, <em>A</em>, in the circuit, I = V/ $R_{eq}$

A = I = 12 V/(5 Ω) = 2.4 A

A = 2.4 A

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