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3 years ago

A generator's maximum output is 220 V. What is the rms potential difference?

1 answer:

3 0

The rms potential difference is defined as the peak value of the potential difference divided by the square root of 2:
$V_{rms} = \frac{V_0}{ \sqrt{2} }$
where
$V_0$ is the peak value of the voltage (the maximum voltage).

The generator in our problem has a maximum voltage of $V_0=220 V$ , so its rms potential difference is
$V_{rms}= \frac{V_0}{ \sqrt{2} }=155.6 V$

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A rock hits a window and stops in 0.15 seconds. The net force on the rock is 58N during the collision. What is the magnitude of

nlexa [21]

Answer:

The change in momentum is $\Delta p = 0.7 \ kg\cdot m \cdot s^{-1}$

Explanation:

From the question we are told that

The time taken for the stone to stop is $\Delta t = 0.15 \ seconds$

The net force on the rock is $F = 58 \ N$

The impulse of the rock can be mathematically represented as

$I = F * \Delta t$

Substituting values

$I = 58 * 0.15$

$I = 0.7\ kg * m * s^{-1}$

Now impulse is defined as the rate at which momentum change

Hence the change in momentum $\Delta p$ of the rock is equal to the impulse of the rock

$\Delta p = I = 0.7 \ kg\cdot m \cdot s^{-1}$

7 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

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3 years ago

How do I do this question?

Mkey [24]

a. 27

after that all you need to do is use the work formula as you will be able to acheive results

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3 years ago

What is the answer of A and B question 2

jok3333 [9.3K]

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2 years ago

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tankabanditka [31]

Answer:28m/s

Explanation:I got one point so yeah hope this help

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2 years ago