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vazorg [7]
3 years ago
7

A balloon rubbed against denim gains a charge of −7.0 µc. what is the electric force between the balloon and the denim when the

two are separated by a distance of 3.0 cm? (assume that the charges are located at a point.) the value of the coulomb constant is 8.98755 × 109 n · m 2 /c 2 . include direction of the force. answer in units of n.
Physics
1 answer:
zubka84 [21]3 years ago
4 0
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
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After a water had boiled, the temperature of the water decreases by 22degrees. The mass of the water in the kettle is 0.50kg.The
kifflom [539]

<u>Answer</u>

46,200 J

The energy of given out when a substance is losing heat is given by:

H = mcΔθ

Where m is the mass of the substance,

            c is the specific heat capacity of the substance and

           Δθ is the temperature change.

H = 0.50 × 22 4200

  = 46,200 Joules.

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4 years ago
Raising the temperature of a gas will increase its pressure if the volume of the gas
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That would be C .............
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3 years ago
A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas
leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₁² v₂

 v_1= \dfrac{r_1^2}{r_2^2} v_2

 v_1= \dfrac{1.4^2}{2.8^2} v_2

 v_1= 0.25 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)

 \Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)

 v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}

 v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}

       v₂ = 1.97 m/s

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Q = A₂ V₂

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5 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
3 years ago
Which ion channels mediate the falling phase of an action potontial?
White raven [17]

Answer:

Voltage-gated K+ channels

4 0
3 years ago
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