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QveST [7]
4 years ago
12

An unknown additional charge q3q3q_3 is now placed at point B, located at coordinates (0 mm, 15.0 mm ). Find the magnitude and s

ign of q3q3q_3 needed to make the total electric field at point A equal to zero.

Physics
1 answer:
Yanka [14]4 years ago
7 0

Answer:

0.3nanocouloumb

Explanation:

Pls see attached file

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A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)
STALIN [3.7K]

The car traverses a distance x after time t according to

x=\dfrac12at^2

where a is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity v after 2 s of travel is given by

v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}

which makes C the most likely answer.

3 0
4 years ago
Coil the wire around the nail. Connect one end of the wire to the positive battery terminal. Connect the other end of the wire t
Andrej [43]
Static i think 
not sure
hope it helps
5 0
3 years ago
Two engineering students, John with a weight of 92 kg and Mary with a weight of 46 kg, are 30 m apart. Suppose each has a 0.04%
Hoochie [10]

Answer:

F = 6.27 x 10 ¹⁹ N

Explanation:

Given

m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol

q₁ = % * [m * N * A * e / M ]  

q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₁ = 3.54 x 10⁶ C

q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₂ = 1.773 x 10⁶ C

Now to determine the electrostatic force con use the equation

F = K * q₁ * q₂ / d²

K = 8.99 x 10 ⁹

F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²

F = 6.27 x 10 ¹⁹ N

3 0
3 years ago
The loudness l of a sound, measured in decibels, is given by l=10log10r, where r is the sound's relative intensity. suppose one
marishachu [46]

Answer

given,                              

I is the loudness of sound

I = 10 Log₁₀ r                  

r is relative intensity                    

at when relative intensity is 10⁶        

I = 60 dB                                                  

how much louder when 100 people would be talking together

I = 10 Log₁₀ r                

I = 10 Log₁₀ (10⁶ x 100)  

I = 10 Log₁₀ (10⁸)                

I = 80 dB                      

hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.

5 0
4 years ago
A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an
nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0
3 years ago
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