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stich3 [128]
3 years ago
11

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail

. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?
Physics
1 answer:
Alik [6]3 years ago
4 0

Answer:

1.34\cdot 10^{-16} C

Explanation:

The strength of the electric field produced by a charge Q is given by

E=k\frac{Q}{r^2}

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C

and the fish can detect the electric field at a distance of

r=63.5 cm = 0.635 m

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C

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Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
The magnitude of the Normal Force on a
lbvjy [14]

Answer:

5. Is greater than mg, always

Explanation:

If the cone has an inclination of angle β, the sum of forces will be:

x-axis (centripetal axis):

N*sin β = m*ax  where ax is the centripetal acceleration

y-axis:

N*cos β - m*g = m*ay   where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.

Solving for N:

N = \frac{m*g + m*ay}{cos \beta }

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.

If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.

7 0
3 years ago
Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v
bazaltina [42]

Answer:

1. t_2 = 2t_1

2. t_2 = t_1\sqrt{2}

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

a_1 = 2a_2

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

t = v / a_1

Now that acceleration is halved:

t = \frac{v}{2a_2}

\frac{v}{a_2} = 2t

You would need to push for twice amount of time t_2 = 2t_1

2. The distance traveled by the puck is as the following equation:

d = at^2

So if the acceleration is halved while maintaining the same d:

\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}

As d_1 = d_2, then d_1/d_2 = 1. Also a_1 = 2a_2

1 = \frac{2a_2t_1^2}{a_2t_2^2}

t_2^2 = 2t_1^2

t_2 = t_1\sqrt{2}\approx 1.14t_1

So t increased by 1.14

7 0
3 years ago
A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo
horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

s=r_{2}-r_{1}

Where, r_{1} = initial position

r_{2} = final position

Put the value into the formula

s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)

s= -6.80i+6.20j-0.1k

(a). We need to calculate the work done on the object

Using formula of work done

W=F\cdot s

Put the value into the formula

W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)

W=-13.6+55.8-0.53

W=41.67\ J

(b). We need to calculate the average power due to the force during that interval

Using formula of power

P=\dfrac{W}{t}

Where, P = power

W = work

t = time

Put the value into the formula

P=\dfrac{41.67}{5.40}

P=7.71\ Watt

(c). We need to calculate the angle between vectors

Using formula of angle

\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})

Put the value into the formula

\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})

\theta=79.7^{\circ}

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

7 0
3 years ago
Need help ??? Please
Yanka [14]
Is there information in the previous question which relates to this one?
8 0
3 years ago
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