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posledela
3 years ago
5

Two point charges carrying charges q1 = +3 μC and q2 = −4 μC are fixed on the x − y plane at positions (x1 = 3.5 cm, y1 = 0.5 cm

) and (x2 = −2 cm, y2 = 1.5 cm) respectively.
a) Find the magnitude and direction of the electrostatic force on charge q2.
b) Where in the x − y plane can you place a third charge q3 = 4 μC, such that the net electrostatic force on charge q2 is zero?

Physics
1 answer:
Stells [14]3 years ago
3 0

Answer:

Explanation:

The concept of coulomb's law was applied to analyze the problem and other mathematical application as is shown in the attached file.

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There is a limit to how long your neck can be. If your neck were too long, no blood would reach your brain! What is the maximum
spayn [35]

Answer:

The maximum height a person's brain could be above his heart is: 1.28 meter.

Explanation:

We need to know what is the normal blood pressure ours hearts so there is a rate: 120/80 (mmHg) and the average will be: 100 (mmHg) and using the Pascal law that relate pressure, density, gravity and height like:P_{2} = pgh_{1} - pgh_{2} + P_{1}, where P is pressure, p is density, g is the gravity acceleration and h is the height. Now we can find the height and delta of pressure will be: P2-P1 = 100 (mmHg), knowing that 1(mmHg) is equal to 133 Pa, we can do the convertion to 13332.2 (Pa), now because the units of Pascal are Newton/(meter^2). Then we solve the formula to get the height: \frac{P2-P1}{pg} =h so we get:\frac{13332.2}{(1060*9,81)}=Height=1.28(meters).

5 0
3 years ago
The New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.20 HzHz , and the
kkurt [141]

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm

Explanation:

Let the total side to side motion be 2A. Where A is maximum acceleration.

Now, we know know that equation for maximum acceleration is;

A = α(max) / [(2πf)^(2)]

So 2A = 2[α(max) / [(2πf)^(2)] ]

α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.

Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm

3 0
3 years ago
Could I get help on this question please . My parents won’t help me /:
vovikov84 [41]

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

From the question given above, the following data were obtained:

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

3 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
NEED ANSWER ASAP!!! Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket.
Schach [20]

a. The free-body diagram for the glass when it is at the top of the circle is attached below.

b. The equation for the net force on the glass at the top of the circle in terms of w, Fn, m, v, and r is mg x g + N -  mg x Vtop² /R =0

c. The glass will fall out of the bucket if the normal force between the glass and bucket equals zero. The speed with which she spin the bucket to prevent this from happening is 3.83 m/s.

d. The string will break if the tension on it is more than 100 N. The range of speeds can  prevent the string from breaking is 3.83< Vtop<4.99 m/s

<h3>What is Net force?</h3>

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given, Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket. She spins the bucket in a vertical circle of radius 1.5 m. She must swing the bucket to keep the glass from falling out.

a. The free body diagram of the bucket and glass is attached below.

b. Bucket will undergo centrifugal force

Fb = mVtop² /R

From the equilibrium of forces, we have

For bucket,

T +mb xg - N =  mb x Vtop² /R..............(1)

For glass,

mg x g + N =  mg x Vtop² /R..............(2)

Thus, this is the net force equation on the glass.

c. On adding both the equations. we have

T + (mb + mg) xg = (mb + mg) Vtop² /R

Substituting the values, T = 0 and from the question, we get

0 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 3.83 m/s

Thus, the speed of spin to prevent glass from falling out is 3.83 m/s

d. The string will break if the tension on it is more than 100 N

100 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 4.99 m/s

Thus, the range of velocity is  3.83< Vtop<4.99 m/s

Learn more about net force.

brainly.com/question/18031889

#SPJ1

8 0
2 years ago
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