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mars1129 [50]
3 years ago
7

Hey guys.. what’s 1,893,9379 x 0

Physics
1 answer:
MakcuM [25]3 years ago
4 0
<h2>Hello there, to answer your question...</h2><h2 /><h2>1,893,9379 x 0 = 0</h2><h2 /><h2>Anything times zero is always 0</h2><h2 /><h2>For example</h2><h2>1x0=0</h2><h2>4x0=0</h2><h2>12x0=0</h2><h2>48x0=0</h2><h3>I hope this helps :)</h3>
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Can anyone tell me what's the base quantities for Force, Pressure and Charge?​
mr Goodwill [35]

Force, pressure, and charge are all what are called <em>derived units</em>. They come from algebraic combinations of <em>base units</em>, measures of things like length, time, temperature, mass, and current. <em>Speed, </em>for instance, is a derived unit, since it's a combination of length and time in the form [speed] = [length] / [time] (miles per hour, meters per second, etc.)

Force is defined with Newton's equation F = ma, where m is an object's mass and a is its acceleration. It's unit is kg·m/s², which scientists have called a <em>Newton</em>. (Example: They used <em>9 Newtons</em> of force)

Pressure is force applied over an area, defined by the equation P = F/A. We can derive its from Newtons to get a unit of N/m², a unit scientists call the <em>Pascal</em>. (Example: Applying <em>100 Pascals </em>of pressure)

Finally, charge is given by the equation Q = It, where I is the current flowing through an object and t is how long that current flows through. It has a unit of A·s (ampere-seconds), but scientist call this unit a Coulomb. (Example: 20 <em>Coulombs</em> of charge)

4 0
3 years ago
8)
marishachu [46]

Answer:

The box will experience an acceleration.

Explanation:

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6 0
3 years ago
The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou
Cerrena [4.2K]

Answer:

a) I = 19.799\,kg\cdot m^{2}, b) T = -3.405\,N\cdot m, c) n_{T} \approx 54.842\,rev

Explanation:

a) The net torque is:

T = I\cdot \alpha

Let assume a constant angular acceleration, which is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}

\alpha = 1.793\,\frac{rad}{s^{2}}

The moment of inertia of the wheel is:

I = \frac{T}{\alpha}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

b) The deceleration of the wheel is due to the friction force. The deceleration is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}

\alpha = - 0.172\,\frac{rad}{s^{2}}

The magnitude of the torque due to friction:

T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )

T = -3.405\,N\cdot m

c) The total angular displacement is:

\theta_{T} = \theta_{1} + \theta_{2}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\theta_{T} = 344.580\,rad

The total number of revolutions of the wheel is:

n_{T} = \frac{\theta_{T}}{2\pi}

n_{T} = \frac{344.580\,rad}{2\pi}

n_{T} \approx 54.842\,rev

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3 years ago
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Mnenie [13.5K]

Answer: C both A and B

Explanation:

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The primary functions of the transmission includes transmitting of the engines crankshaft into a more usable rotational speed and to multiply the torque of the engine's output. Hence,

Technician B was also right.

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3 years ago
When an object is in motion due to only the force of gravity?
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When it falls though a vaccum,it is subjected to only one external force,the gravitational force.
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