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2 years ago

Ronald likes to use his erector set more than anything else.

2 answers:

5 0

C: Foreclosure. People in identity foreclosure have committed to an identity too soon. Often they have simply adopted the identity of a parent, close relative or respected friend.

MAVERICK [17]2 years ago

5 0

Answer:

C. forecloser

Explanation: In forecloser identity people wants their identity as soon as possible. Generally they simply adopt the identity of persons close to their daily life such as father, brother, friends or relatives.In the given case Ronald is also trying to get identified soon and wants be like his father and brother in the future, so he is expressing a forecloser identity

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Consider a capacitor made of two rectangular metal plates of length and width , with a very small gap between the plates. There

mezya [45]

Answer:

F= σ² L² /2ε₀

F = (L² ε₀/4π) ΔV² / r⁴

Explanation:

a) For this exercise we can use Coulomb's law

F = - k Q² / r²

where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates

Capacitance is defined by

C = Q / ΔV

Q = C ΔV

also the capacitance for a parallel plate capacitor is related to its shape

C = ε₀ A / r

we substitute

Q = ε₀ A ΔV / r

we substitute in the force equation

F = k (ε₀ A ΔV / r)² / r²

k = 1 / 4πε₀

F = ε₀ /4π L² ΔV² / r⁴4

F = L² ΔV² ε₀/ (4π r⁴)

F = (L² ε₀/4π) ΔV² / r⁴

b) Another way to solve the exercise is to use the relationship between the force and the electric field

F = q E

where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface

Ф = ∫E .dA = $q_{int}$ / ε₀

the plate have two side

2E A = q_{int} / ε₀

E = σ / 2ε₀

σ = q_{int} / A

substituting in force

F = q σ / 2ε₀

the charge total on the other plate is

q = σ A

q = σ L²

F= σ² L² /2ε₀

4 0

2 years ago

A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it i

il63 [147K]

Answer:

a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to

This can be calculated by Gauss' Law.

A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.

b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>

8 0

2 years ago

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

ehidna [41]

Answer

The Value of r = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

$R_T = R_1 +R_2$

And from Ohm law

$I_s =\frac{ V}{R_T}$

$I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel is

$R_T = \frac{1}{R_1} +\frac{1}{R_2}$

$= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

=> $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

Dividing equation 2 with equation 1

=> $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

$10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3

$10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

$r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

a = 1 b = -8 c =1

$= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

$r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

$r = \ 7.87\ or \ r \ = \ 0.127$

Now r = 0.127 because it is the least value among the obtained values

4 0

2 years ago

A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many

jekas [21]

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

= 57units

5 0

2 years ago

The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo

agasfer [191]

As we know by Doppler's effect of light we have

$\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

$3.01 \times 10^{-3} = \frac{v}{c}$

here we have

$v = 904.3 km/s$

since wavelength is increased so we can say that it is moving away

so correct answer is

1- 904.3 km/s away from the Earth

3 0

2 years ago