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3 years ago

Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?

1 answer:

8 0

T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,

Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.

</span>

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otez555 [7]

Answer:

The mass of the receiver is 68.862 kg

Explanation:

Given;

let mass of the receiver be "m₁"

initial velocity of the receiver, u₁ = 0

mass of the tackler, m₂ = 156 kg

initial velocity of the tackler, u₂ = 6.89 m/s

their final velocity, v = 4.78 m/s

Apply the principle of conservation of linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(m₁ x 0) + (156 x 6.89) = 4.78(m₁ + 156)

0 + 1074.84 = 4.78 (m₁ + 156)

1074.84 = 4.78 (m₁ + 156)

m₁ + 156 = 1074.84 / 4.78

m₁ + 156 = 224.862

m₁ = 224.862 - 156

m₁ = 68.862 kg

Therefore, the mass of the receiver is 68.862 kg

5 0

3 years ago

A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

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1 year ago

СРОЧНО ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!!!!!!<br>

Deffense [45]

Боже, как это сложно! Ну ладно.

Между прочим это ты сам должен делать, а то не куда не поступишь!

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3 years ago

A rain drop of radius 0.3mm has terminal velocity 1m/s in air. The viscosity of air is 18X10-5 poise, then viscous force is

dimaraw [331]

Answer:

F = 1.0178 × 10^(-2) dyne

Explanation:

From stokes law, the viscous force also known as drag force on rain drop is given by the formula;

F = 6πηrv

Where;

η is viscosity

r is radius

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We are given;

η = 18 × 10^(-5) poise

r = 0.3 mm = 0.03 cm

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Thus;

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3 years ago

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Answer:

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Explanation:

see the attached file

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4 years ago