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3 years ago

If you take 3s to do 72J of work, how much power did you exert?

Physics

1 answer:

Svetradugi [14.3K]3 years ago

7 0

Answer:

The answer to your question is Power = 24 Watts

Explanation:

Data

time = 3 s

Work = 72 J

Power = ?

Power is defined as the rate at which a work is done.

Formula

Power = Work / time

-Substitution

Power = 72/3

-Simplification

Power = 24 Watts

-Conclusion

The Power exerts by 72 J of work in 3 seconds is 24 watts.

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A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar

BlackZzzverrR [31]

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

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3 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

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3 years ago

Coal burns in a furnace, producing light and heat. This reaction is

olasank [31]

The reaction is Exothermic

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3 years ago

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In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles

alexandr402 [8]

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4 x 10⁻¹⁵ m

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3 years ago

The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t

REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

From the question we are told that

The initial Volume is $Vi = 0.160 L$

The final volume is $V_f = 0.510L$

The external pressure is $P = 5.00 \ atm$

Generally the change in volume is

$\Delta V = V_f - V_i$

Substituting values we have

$\Delta V = 0.510 -0.160$

$= 0.350L$

Generally workdone is mathematically represented as

$W = -P \Delta V$

W is negative because the working is done on the environment by the system which is indicated by volume increase

Substituting values

$W = - 5* 0.350$

$= -1.75 \ L \ \cdot atm$

Now $1 \ L \cdot atm = 101.3J$

Therefore $W = -1.75* 101.3$

$= -177.275J$

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3 years ago

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