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kozerog [31]
3 years ago
9

An electroscope is a simple device consisting of a metal ball that is attached by a conductor to two thin leaves of metal foil p

rotected from air disturbances in a jar, as shown. When the ball is touched by a charged body, the leaves that normally hang straight down spread apart. Why? (Electroscopes ate useful not only as charge detectors but also for measuring the quantity of charge: the more charge transferred to the ball, the more the leaves diverge.)
Physics
1 answer:
baherus [9]3 years ago
3 0

Answer:

the electroscope separate  by the presence of charge carriers

Explanation:

Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where

                  Fe - Tx = 0

                  Fe = Tx

In summary, the electroscope separate its leaves by the presence of charge carriers

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A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
A 1500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed
IgorC [24]

Answer:

The direction and magnitude of velocity is 38.65° and 12.005 m/s

Explanation:

Given that,

Mass of car = 1500 kg

Speed of car = 25.0 m/s

Mass of van = 2500 kg

Speed of van =20.0 m/s

We need to calculate the velocity

Using conservation of energy

m_{c}u_{i}+m_{v}u_{i}=(m_{c}+m_{v})v_{f}

1500(25i+0j)+2500(0+20j)=4000(v_{f})

v_{f}=\dfrac{1500\times25}{4000}i+\dfrac{1500\times20}{4000}j

v_{f}=9.375 i+7.5 j

The magnitude of velocity

|v_{f}|=\sqrt{(9.375)^2+(7.5)^2}

|v_{f}|=12.005\ m/s

We need to calculate the direction

\tan\theta=\dfrac{coefficient\ of\ j}{coefficient\ of\ i}

\tan\theta=\dfrac{7.5}{9.375}

\theta=\tan^{-1}0.8

\theta=38.65^{\circ}

Hence, The direction and magnitude of velocity is 38.65° and 12.005 m/s.

3 0
4 years ago
A bucket weighing 5 lbs is lifted at a constant rate from the bottom of a 100 ft well by a rope which weighs 5 lbs. The bucket h
8_murik_8 [283]

Answer:

W_{bucket} = 24934.85\,lbf\cdot ft

Explanation:

The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

W_{bucket} = U_{g,B, bucket} - U_{g,A,bucket} +U_{g, B, water}-U_{g,A,water} +U_{g, B, rope} -U_{g,A,rope}

W_{bucket} = (5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-0\,ft) + (25\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft) - (30\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (0\,ft)+(5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-50\,ft)

W_{bucket} = 24934.85\,lbf\cdot ft

6 0
4 years ago
A 1300-kg car moving on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to
slega [8]

The stiffness constant of the spring is 68,290.3 N/m

<h3> Stiffness constant of the spring</h3>

Apply the principle of conservation of energy;

U = K.E

¹/₂kx² = ¹/₂mv²

kx² = mv²

k = mv²/x²

where;

  • m is mass
  • v is speed = 60 km/h = 16.67 m/s
  • x is the distance

k = (1300 x 16.67²)/(2.3²)

k = 68,290.3 N/m

Thus, the stiffness constant of the spring is 68,290.3 N/m.

Learn more about stiffness constant here: brainly.com/question/1685393

#SPJ1

7 0
3 years ago
A multiparous client presents to the labor and delivery area in active labor. The initial vaginal examination reveals that the c
balandron [24]

Answer:

correct answer is Precipitous vaginal delivery

Explanation:

given data

cervix dilated = 4 cm

effaced = 100%

delivery  = 5 minutes later

solution

correct answer is Precipitous vaginal delivery because precipitous take delivery time less than = 3 hours

A multipara progress at rate 1.5 cm of dilation per hour

and it is progress for 10 cm for the deliver and birth averages approx 20 minute

so here correct answer is Precipitous vaginal delivery

4 0
3 years ago
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