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natima [27]
3 years ago
5

The SI unit of force is the

Physics
1 answer:
dexar [7]3 years ago
6 0
The SI Measuring Unit for force is the Newton
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If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 n/c, the air breaks down and a spark forms. for a two-di
Vlad1618 [11]

Answer: 39.8 μC

Explanation:

The magnitude of the electric field generated by a capacitor is given by:

E = \frac{V}{d}

d is the distance between the plates.

For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

C =\frac{\epsilon_o A }{d}

where A is the area of the plate and ε₀ is the absolute permittivity.

substituting, we get

E = \frac{Q}{\epsilon_o A}

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.

radius of the plates of the capacitor, r = 69 cm = 0.69 m

Area of the plates, A = πr² = 1.5 m²

Thus, the maximum charge that can be placed on disks without a spark is:

Q = E×ε₀×A

⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.

8 0
3 years ago
Which one of these is an exanple of precepation?
Leokris [45]

Answer:

Snow

Explanation:

Precipitation is the formation of a solid after being a liquid. Snow, which is a solid, forms from water, a liquid.

8 0
2 years ago
Read 2 more answers
1.Sobre a queda-livre,assinale V para verdadeiro e F para falso
lisov135 [29]

Answer:pelo o que eu sei é ..

V

V

V

F

F

F

Explanation:

8 0
3 years ago
Please help me I really need it
tensa zangetsu [6.8K]

Answer:

A 1.5 metres per second squared B 0 C 6 0.5

b) 9m

Explanation:

a= v-a/t

a= acceleration measured in metres per second squared

v= final speed measured in metres per second

u= initial speed measured in metres per second

t= time measured in second

A

a=v-a/t

a= 3-0/2

a= 1.5 metres per second squared

Distance in a velocity time graph = area under

To calculate the distance in the first 4 seconds

You divide the graph into triangles and rectangles

Area A (triangle )= 1/2 × base × height

=1÷2 ×2×3

= 3m

Area B (rectangle)= L×B

=2×3

=6m

Area A + Area B= 3+6

= 9m

4 0
3 years ago
Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What
hjlf

To solve this problem we will use the given expression and derive it in order to find the algebraic expressions of velocity and position. These equations will be similar to those already known in the cinematic movement but will be subject to the previously given function. We start deriving the equation for velocity

a = 2t-6

\frac{dv}{dt} = 2t-6

Integrate acceleration equation

\int dv = (2t-6)dt

v = 2(\frac{t^2}{2})-6t+C_1

v=t^2-6t+C_1

At t = 0, v = 0

Replacing,

0 = 0^2-6*0+C_1

Therefore the value of the first Constant is

C_1 = 0

The expression can be escribed as,

v = t^2-6t

Calculate the velocity after 6s,

v=t^2-6t

v = 6^2-6*6

v = 0m/s

Now using the same expression we can derive the equation for distance

v = t^2-6t

\frac{dx}{dt} =t^2-6t

\int dx = \int (t^2-6t)dt

x = \frac{t^3}{3}-6\frac{t^2}{2}+C_2

At t=0, x=0

0 = \frac{0^3}{3}-6(\frac{0^2}{2})+C_2

Therefore the value of the second constant is

C_2 = 0

x = \frac{t^3}{3}-\frac{6t^2}{2}

Calculate the distance traveled after 11 s

At  t=11s

x = \frac{11^3}{3}-6(\frac{11^2}{2})

x = 80.667m

6 0
3 years ago
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