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3 years ago

14

a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the

postive charge

1 answer:

Alika [10]3 years ago

5 0

Answer:

The force is 86.5×10^9 N towards the negative charge (to the right)

Explanation:

The electrostatic force on the charges is given by Coulomb's law;

F= Kq1q2/r^2

This an inverse square law.

F= electrostatic force on the charges

K= constant of Coulomb's law

q1 and q2= magnitude of the charges

Since K= 9.0×10^9Nm^2C^2

F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56

F= 86.5×10^9 N

The force is 86.5×10^9 N towards the negative charge.

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Build a filter.
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3 years ago

Richard Julius once made a model plane that could travel a max speed of 110 m/s. Suppose the plane was held in a circular path b

hjlf

Answer:

85.8 m/s

Explanation:

We know that the length of the circular path, L the plane travels is

L = rθ where r = radius of path and θ = angle covered

Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt

where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path

Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,

θ = θ₀ + ωt = 0 + ωt = ωt

So, v = rdθ/dt + θdr/dt

v = rω + ωtdr/dt

v = (r + tdr/dt)ω

v = (r + tdr/dt)v'/r

v = v' + tv'/r(dr/dt)

v = v'[1 + t(dr/dt)/r]

Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)

So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]

v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]

v = 110 m/s[1 + 11.0 s(-0.02/s)]

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3 years ago

A man wishes to row the shortest possible distance from north to south across a river which is flowing at 2 km/hr from the east.

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From the diagram we have that

$sin\theta = \frac{2}{4}$

$\theta = sin^{-1} (\frac{1}{2})$

$\theta = 30\°$

Therefore the direction is 30° from east of south

8 0

3 years ago

Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is t

den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

$F\propto \dfrac{1}{r^2}$

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N

We need to find the electric force between charges if the new separation of 0.25 cm. So,

$\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N$

So, the new force is 64 N if the separation between charges is 64 N.

7 0

3 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

6 0

3 years ago