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Sergeu [11.5K]
3 years ago
11

What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?

Physics
1 answer:
ratelena [41]3 years ago
4 0
= 80 N/40 kg
= 2 m/s 2
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Two objects of the same mass travel in opposite directions along a horizontal surface. Object X has a speed of 5ms and object Y
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Explanation:

Because theyre heading opposite directions

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We all depend on electricity. Most electricity is created by electromagnetic generators at large power plants and distributed th
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1) not so long (maybe an hour or two)

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4) solar panels can be used to draw power from incident sun rays, this power can be stored in an inverter for future use in case of a power outage.

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8 0
3 years ago
a net force of 219 N is exterted on a rock. the rock has an acceleration of 3m/s^2 due to this force. what is the mass of the ro
Sonbull [250]

Answer:

<h2>73 kg</h2>

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m =  \frac{f}{a}  \\

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From the question we have

m =  \frac{219}{3}  \\

We have the final answer as

<h3>73 kg</h3>

Hope this helps you

6 0
3 years ago
The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G
Charra [1.4K]

Answer:

Atomic Size and Mass:

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7 0
3 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
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