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2 years ago

11

A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant

is 10^6N/M.neglecting friction,how much work does the truck do on the spring?

1 answer:

mixas84 [53]2 years ago

3 0

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

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A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows

jeka57 [31]

Answer:

(a) The impedance in the circuit is $Z=183.33\Omega$ .

(b)The resistance is $R=38.89\Omega$ .

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

$Z=\frac{V_rms}{I_rms}$

Here, $V_rms$ is the rms voltage and $I_rms$ is the rms current.

Put $V_rms=110 V$ and $I_rms=0.600 A$ .

$Z=\frac{110}{0.600}$

$Z=183.33\Omega$

Therefore, the impedance in the circuit is $Z=183.33\Omega$ .

(b)

The expression for the average power is as follows;

$P_{a}=I_{rms}^{2}R$

Here, $P_{a}$ is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

$R=\frac{P_{a}}{I_{rms}^{2}}$

Put $P_{a}=14W$ and

$R=\frac{14}{{0.600}^{2}}$

$R=38.89\Omega$

Therefore, the resistance is $R=38.89\Omega$ .

(c)

The expression for the impedance is as follows;

$Z^{2}=R^{2}+X_{L}^{2}$

Here, $X_{L}$ is the inductive reactance.

Put $Z=183.33\Omega$ and $R=38.89\Omega$ .

$(183.33)^{2}=(38.89)^{2}+X_{L}^{2}$

$X_{L}=179.16\Omega$

The expression for the inductive reactance in terms of frequency is as follows;

$X_{L}=2\pi fL$

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

$L=\frac{X_{L}}{2\pi f}$

Put $X_{L}=179.16\Omega$ and f=50Hz.

$L=\frac{179.16}{2\pi (50)}$

L=0.57 H

Therefore, the inuctance is 0.57 H.

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