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3 years ago

How do you increase capacitance in a capacitor?

1 answer:

4 0

If you want to increase the Capacitance of Parallele Plate Capacitor then increase the surface area, reduce the separation between the plate and use a dielectric material in between the plate which have higher dielectric breakdown strength.

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A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block c

schepotkina [342]

After impact velocity = 14.968 ft/s

Weight and mass of Bullet and wooden block:

Bullet: w = 1oz = 1/16 lb m = 0.001941 lb

wooden block : W = 5lb M = 0.15528 lb

velocity of block and bullet immediately after impact:

Σmv1 + ΣImp = mv2

Resolving vertical component

( m× v₀cos30⁰) + 0 = ( m+M) v'

v' = ( m× v₀cos30⁰)/ (m+M)

v' = 14.968 ft/s

Horizontal and vertical component of the impulse exerted by block on the bullet:

Here we will apply the principle of impulse and momentum.

Horizontal component:

-mv₀ cos30⁰ + RxΔt =0

RxΔt = mv₀sin30⁰

= 0.001941 × 1400sin30⁰

RxΔt = 1.3587 lb.s

Vertical component:

-mv₀cos30⁰ + RyΔt = -mv'

RyΔt = m( v₀cos30⁰-v')

RyΔt = 0.001941(1400cos30⁰ - 14.968)

= 2.32 lb.s

Learn more about impact here:

brainly.com/question/15008937

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8 0

1 year ago

PLEASE HELP ASAP!!!!!!!!!!

Marianna [84]

Hey,

I think the answer's are 1,3

Hope this helpss

~Girlygir101~

7 0

3 years ago

A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: $E_k=\frac{1}{2}mv^2$

2. gravitational potential energy of mass m, grav. acc. g and height h: $E_g=mgh$

3. potential energy in a spring with spring constant k and displacement from equilibrium x: $E_s=\frac{1}{2}kx^2$

Calculating x:

$\frac{1}{2}mv_a^2=\frac{1}{2}kx^2$

$x=\sqrt{\frac{m}{k}}v_a$

Calculating the speed:

$\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}$

$h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R$

$\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R$

Solving for $v_b$ :

$v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}$

7 0

3 years ago

For a steady two-dimensional flow, identify the boundary layer approximations.

Georgia [21]

Answer:

The velocity component in the flow direction is much larger than that in the normal direction ( A )
The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )

Explanation:

For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction

assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that

Vy << Vx

4 0

3 years ago

If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?

pishuonlain [190]

The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net - ( 1 )

Step 1:

The net electric field value is given as:

E_net = E₁ cos Φ + E₂ cos Φ

= 2E₁ cos Φ -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod

respectively.

Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r) - ( 3 )

where, k is Coulomb's force constant

λ is linear charge density

r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

= √1.08

= 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

= k [(Q/x)/r]

= k [ Q/xr ]

= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

= 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:

E₂ = k [ Q/xr ]

= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

= 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

= tan⁻¹ ( 1 )

= π/4

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

= 2(1.803×10⁴ N/C) (0.5)

= 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

= 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Learn more about electric force here:

<u>brainly.com/question/1634182</u>

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8 0

2 years ago