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Burka [1]
3 years ago
13

When a positively charged conductor touches a neutral conductor, the neutral conductor will:

Physics
1 answer:
frutty [35]3 years ago
6 0

Answer:

Lose electrons

Explanation:

When a positively charged conductor touches a neutral conductor, the neutral conductor will lose electrons. Only electrons can move from one conductor to another, so if the neutral conductor ended up with a positive charge it means it lost electrons. The conductor touching and the neutral conductor both end up being charged positively.

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A steel ball of mass 0.1 kg falls from a height of 1.8metres
murzikaleks [220]
And what about it?????????
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The slope of a linear velocity-time graph tells us the _____ of the object.
Andrew [12]
Speed of the object !!!!
4 0
3 years ago
John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and
EleoNora [17]

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

Therefore, the keys will make it back to John.

7 0
3 years ago
‼️‼️ Please help, urgent ‼️‼️ (check photo)
Alex787 [66]

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}

The total energy at the point h=0m is:

E_{tot}=E_{kin}+E_{pot}+Work=0+0+ Work\\E_{tot} =F_{friction}\Delta h+\frac{1}{2}k (\Delta h)^2=17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2

The two Energy values are to be equal (by law of energy conservation), which allows us to determine the only unknown, namely the force constant k:

17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2 = 55200 \,kg\frac{m^2}{s^2}\\k = \frac{55200-34000}{2}\,\frac{kg}{s^2}=10600\frac{kg}{s^2}

5 0
3 years ago
A force of 900 N is applied to a 9kg object. How fast will it accelerate?
saul85 [17]

Answer:  15 m/s2

Explanation: I hope this helps or right because I learned this a few months ago

7 0
1 year ago
Read 2 more answers
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