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HACTEHA [7]
3 years ago
8

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

d q2 = -4.0 µc, x2 = -2.0 cm, y2 = 1.5 cm. (a) find the magnitude and direction of the electrostatic force on q2
Physics
1 answer:
Readme [11.4K]3 years ago
7 0
1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
where k_e=8.99\cdot10^{9} Nm^2C^{-2} is the Coulomb's constant.
Substituting numbers, we get 
F=8.99\cdot10^{9} Nm^2C^{-2}  \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
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Write a speech about bringing back fidget toys because they're banned in school
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The voltage between two parallel plates separated by a distance of 3. 0 cm is 120 v. The electric field between the plates is?
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The electric field between plates is 4000V/m.

An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.

The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.

The voltage between points A and B is

V=E.d

E =V/d  (uniform E- field only)  

where  d is the distance from A to B, or the distance between the plates.

Given:

distance d = 3 cm

voltage V = 120 V

Electric field E = V/d

                     E = 120 V / 3cm

                     E = 40 V / 1 cm               [ 1 cm = 1/100 m ]

                    E = 4000 V/m.

Learn more about Electric field here:

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6 0
1 year ago
What are applications of zeroth law of thermodynamics?​
Harrizon [31]

Answer:

Applications of zeroth law of thermodynamics:

1. When we get very hot food, we wait to make it normal. In this case, hot food exchanges heat with surrounding and brings equilibrium.

2. We keep things in the fridge and those things come equilibrium with fridge temperature.

3. Temperature measurement with a thermometer or another device.

4. In the HVAC system, sensors or thermostats are used to indicate temperature. It always comes in a thermal equilibrium with room temperature.

5. If you and the swimming pool you’re in are at the same temperature, no heat is flowing from you to it or from it to you (although the possibility is there). You’re in thermal equilibrium.

6 0
2 years ago
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of
Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

3 0
2 years ago
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