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aliya0001 [1]
3 years ago
13

As you move away from a positive charge distribution, the electric field:

Physics
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

E=\dfrac{kq}{r^2}

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

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What's the main function for the human digested system​
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A car's bumper is designed to withstand a 7.20 km/h (2.0 m/s) collision with an immovable object without damage to the body of t
ollegr [7]

Answer:

8512 N

Explanation:

From the work energy theorem we know that: The net work done on a particle equals the change in the particles kinetic energy:

W=\Delta K=K_{f}-K_{i} \\
\\qquad \begin{array}{r}
W=F \cdot d, \Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\
F \cdot d=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}

Where:

-W is the work done by the force.

- F is the force actin on the.

- d is the distance travelled.

- m is the mass of the car.

- v_{f}, v_{i} are the final and the initial velocity of the car

K_{f}, K_{i} are the final and the kinetic energy of the car.

Givens: m=830 \mathrm{~kg}, v_{i}=2 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~km} / \mathrm{h}, d=0.195 \mathrm{~m}

Plugging known information to get:

F \cdot d &=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\
F &=\frac{\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}}{d} \\
&=\frac{0-\frac{1}{2} \times 830 \times 2^{2}}{0.195} \\
&=8.512 \times 10^{3} \\
F &=8.512 \times 10^{3} \mathrm{~N}

8 0
3 years ago
The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If
ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

y=3+0.8x-0.4x^2........(1)

The x component of constant velocity, v_x=5\ m/s

We need to find the resultant velocity at the point (2,3).

Let \dfrac{dx}{dt}=v_x and \dfrac{dy}{dt}=v_y

Differentiating equation (1) wrt t as,

\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}

v_y=0.8\times v_x-0.8x\times v_x

v_y=0.8v_x(1-x)

When x = 2 and v_x=5\ m/s

So,

v_y=0.8\times 5\times (1-2)

v_y=-4\ m/s

Resultant velocity, v=\sqrt{v_x^2+v_y^2}

v=\sqrt{5^2+(-4)^2}

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0
3 years ago
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