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GenaCL600 [577]

3 years ago

14

Tim can type 2000 words at a speed of 50 words per minute bobs typing speed is 80 words per minute how much less time did bob ta

ke than Tim to type 2000 words

1 answer:

Whitepunk [10]3 years ago

8 0

Answer:

15 less mins will be used by Bob

Explanation:

This is because it takes Tim 2000/50= 40mins to type the whole work

While it takes Bob 2000/80= 25 mins

So the difference 40-25= 15mins

Will be 15mins

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Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

3 0

3 years ago

If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ra

igomit [66]

Answer:

$\frac{E}{E_c} =3.125$

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

$E_c=\frac{1}{2} mv^2$

The equivalence between mass and energy established by the theory of relativity is given by:

$E=mc^2$

This formula states that the equivalent energy $E$ can be calculated as the mass $m$ multiplied by the speed of light $c$ squared.

Where $c$ is approximately $3\times 10^{8} m/s$

Hence:

$E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J$

$E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J$

Therefore, the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

$\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125$

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F.r.e.e points :]]]

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Ok? I don’t know what you want me to do though

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