3 years ago

A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:

Physics

1 answer:

alexdok [17]3 years ago

7 0

Answer:

correct me if im wrong

brainlest plsss<333

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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

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Answer:

$h = 5.505\,m$

Explanation:

Let assume that ball begins its movement at a height of zero. The height can be found by energy approach, to be precise, by Principle of Energy Conservation:

$\frac{1}{2}\cdot (0.145\,kg)\cdot (12\,\frac{m}{s} )^{2}= \frac{1}{2}\cdot (0.145\,kg)\cdot (6\,\frac{m}{s} )^{2}+(0.145\,kg)\cdot (9,81\,\frac{m}{s^{2}} )\cdot h$

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When the mass of a body is constant, then its acceleration is directly proportional to the force acting on it. When the force ac

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Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

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