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Pavlova-9 [17]
3 years ago
11

,how do charged objects react???quiet urgent

Physics
2 answers:
son4ous [18]3 years ago
8 0
Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>
galina1969 [7]3 years ago
3 0
Any charged subject can exert the force upon other subjects...I think...I hope this helped you...And I hope you have a good day!!! :) :) :) :) :) :)
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Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ign
iren [92.7K]

plane is flying at an altitude of 70 m

now if an object is dropped from it then time taken by object to drop on ground will be given as

y = v_i* t + \frac{1}{2}at^2

here initial speed in vertical direction must be zero as plane is moving horizontal

given that

y = 70 m

a = 9.8 m/s^2

70 = 0 + \frac{1}{2}*9.8*t^2

t = 3.77 s

now since the plane is moving horizontally with speed v = 44 m/s

so the horizontal distance moved by the object will be

d = v_x * t

d = 44 * 3.77

d = 166.3 m

so the distance moved by the box is 166.3 m

3 0
3 years ago
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Which statement BEST explains why the specific heat of water is higher than the specific heat of most other substances?
solniwko [45]
Due to its polarity and hydrogen bonding water can absorb heat without a significant temperature change.. The high specific heat of water helps regulate the rate at which air changes temperature, which is why the temperature change between seasons is gradual instead of sudden, especially near the oceans.
5 0
3 years ago
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Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d
xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F =   0.0111 \ N

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0
3 years ago
A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t
Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                                P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                               P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                               P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                      = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                  ∴                 K.E = P.E

                                     1/2 mv² = 138.44

                                     1/2 x 25 x v² 138.44

                                            v² = 11.0752

                                             v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

             Tension on string, T = Force acting on the swing, F

                      W=L\int\limits^0_\phi{F} \, d \phi

                             =L\int\limits^0_\phi{mg.sin \phi} \, d \phi

                            = -Lmg[cos\phi]_{42}^{0}

                            = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                            = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0
4 years ago
Any one want to talk i'm so boreed
Sedbober [7]

Answer:

<h3>We can talk. ❤❤</h3>

Explanation:

..........

3 0
3 years ago
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