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Alex777 [14]
3 years ago
9

Air will break down (lose its insulating quality) and sparking will result if the field strength is increased to about 3 × 106 N

/C. What acceleration will an electron experience in such a field? The mass of the electron is 9.109 × 10−31 kg and the fundamental charge is 1.602 × 10−19 C.
Physics
1 answer:
CaHeK987 [17]3 years ago
6 0

Answer:

Explanation:

Given

Electric Field Strength E=3\times 10^6\ N/C

When a charged particle enters the Electric field it Experience a force in the direction of electric field (for positive charge and vice-versa for negative charge)

charge on Electron q=1.6\times 10^{-19}\ C

mass of electron m=9.1\times 10^{-31}\ kg

Force F=q\times E

F=1.6\times 10^{-19}\times 3\times 10^6

F=4.8\times 10^{-13}\ N

acceleration of electron a=\frac{F}{m}

a=\frac{4.8\times 10^{-13}}{9.1\times 10^{-31}}

a=0.527\times 10^{18}\ m/s^2      

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A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc
kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

\huge\boxed{\Sigma \tau = I \alpha}

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}

4 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

4 0
3 years ago
If the forces acting on an system produce a net force, what can we say about these forces
lutik1710 [3]
Net force refers to the (vector) sum of all the forces acting on something. It's a mathematical construction, so it's not a single identifiable force. The forces themselves are real, but the net force is not an actual force.

Hope this helps you out.
8 0
3 years ago
A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?
Lyrx [107]
W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.
3 0
3 years ago
Read 2 more answers
A 500 N weight is hung at the middle of a rope attached to two buildings at the same level. If the breaks in the tension exceed
Lyrx [107]

Not sure what you mean by "breaks in the tension" but I suspect you mean the rope will come apart if the tension in the rope exceeds 1800 N.

In the free body diagram for the 500 N weight, we have a figure Y with the net force equations

• horizontal net force:

∑ F[hor] = T₁ cos(θ) - T₂ cos(θ) = 0

• vertical net force:

∑ F[ver] = T₁ sin(θ) + T₂ sin(θ) - 500 N = 0

From the first equation, it follows that T₁ = T₂, so I'll denote their magnitude by T alone. From the second equation, we have

2 T sin(θ) = 500 N

and if the maximum permissible tension is T = 1800 N, it follows that

sin(θ) = (500 N) / (3600 N)   ⇒   θ = arcsin(5/36) ≈ 7.9°

is the smallest angle the rope can make with the horizontal.

6 0
3 years ago
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