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DanielleElmas [232]
2 years ago
13

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50

Physics
1 answer:
Setler79 [48]2 years ago
6 0

Answer:

very hard others will answer it

Explanation:

hard

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A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one
Fittoniya [83]
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
8 0
3 years ago
Read 2 more answers
Most of the elements (85%) in the periodic table are.
Rasek [7]
Most of the elements are metals
4 0
3 years ago
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A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns
Sveta_85 [38]

Answer:

182.28 W

Explanation:

Here ,

m = 7.30 Kg

distance , d= 28.0 m

time , t = 11.0 s

average power supplied = change in potential energy/time

average power supplied = m×g×d/time

average power supplied = 7.30×9.81×28/11

average power supplied = 182.28 W

the average power supplied is  182.28 W

6 0
2 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
An electromagnet is a device in which moving electric charges (current) in a coil of wire create a magnet. What’s one advantage
yan [13]
I think the correct answer from the choices listed above is option D. One advantage of using electromagnets in devices would be that electromagnets can <span>easily be turned on and off. Hope this answers the question. Have a nice day.</span>
7 0
3 years ago
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