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Evgesh-ka [11]
2 years ago
13

How does electricity affect objects that are not in contact with one another?

Physics
1 answer:
Elan Coil [88]2 years ago
5 0

Electrostatic forces are non-contact forces; they pull or push on objects without touching them

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A 2 kg object with a weight of 20 N is being pulled up by a rope with a tension of 12N what is the acceleration of the object
son4ous [18]

Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2 (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

7 0
3 years ago
Plz help!
Strike441 [17]
72km/hr = 72000m/3600seconds(we convert km/hr to m/s
= 20m/s
Velocity= 20m/s
In 4seconds,distance covered=20x4= 80m
Therefore,an accident occurs because in 4seconds, the car would have moved past 20m
6 0
3 years ago
our 3.80-kg physics book is placed next to you on the horizontal seat of your car. The coefficient of static friction between th
gtnhenbr [62]

Answer:

Explanation:

Maximum force of friction possible = μmg

= .65 x 3.8 x 9.8

= 24.2 N

u = 72 x 1000 / 60 x 60

= 20 m /s

v² = u² - 2as

a = 20 x 20 / (2 x 30)

= 6.67 m / s²

force acting on it

= 3.8 x 6.67

= 25.346 N

Friction force possible is less .

So friction will not be able to prevent its slippage

It will slip off .

4 0
3 years ago
This was a british philosopher is widely regarded as one of the most influential of enlightenment thinkers and commonly known as
Darya [45]

Is there a question? Because All your doing t explaining a british philosopher to us..

4 0
3 years ago
Read 2 more answers
Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm
Phantasy [73]

Answer:

1.44 x 10⁻⁶ C

Explanation:

q_{1} = charge on one sphere

q_{2} = charge on other sphere

q = Total charge on the two spheres = 40 μC

q_{1}+ q_{2} = q

q_{1}+ q_{2} = 40 x 10⁻⁶

q_{1} = (40 x 10⁻⁶) - q_{2}                                   eq-1

r = distance between the two spheres = 50 cm = 0.50 m

F = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

F = \frac{k q_{1} q_{2}}{r^{2}}

2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}

q_{2} = 1.44 x 10⁻⁶ C

7 0
3 years ago
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