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2 years ago

3. What is the weight of a 462-kg bar?

Physics

1 answer:

FromTheMoon [43]2 years ago

3 0

Answer:

F = M g = 462 kg * 9.80 m/s^2 = 4528 Newtons

It somewhat unfortunate that weight in the English System is measured in pounds (mass in slugs) and in the metric system weight is measured in Newtons (mass in kilograms)

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Nathalie leaves a history classroom and walks 3 meters North to drinking fountain. Then she turns and walks 10 meters south to a

pishuonlain [190]

Answer:

13 meters

Explanation:

Step one:

given

We are told that Nathalie leaves a history classroom and walks 3 meters North

Then travels another 10 meters south to an art classroom.

Required

The total distance.

Step two:

The total distance can be computed by summing up the 3 meter North distance traveled and the 10 meter south distance traveled

Total distance= 3+10= 13meters

7 0

2 years ago

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Anika [276]

Answer:

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force $\times$ distance

$= 600 \times 5$

Calculate the product

$= 3000 \:$ Joule

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

Read 2 more answers

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

2 years ago

4-08 t 1 - / S P = 1 / P = S 1 S= P Def. of Speed

nordsb [41]

Answer:

what do I answer? not enough information

8 0

2 years ago

A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex

Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

$\text{Efficiency of the heat engine} = \eta = 26\% = 0.26$