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3 years ago

A point charge Q is located a short distance from a point charge 3Q, and no

Physics

1 answer:

ruslelena [56]3 years ago

5 0

The other force is also 'F'.

The forces on any two charged particles are opposite and EQUAL, and proportional to the product of the charges.

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Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00

lions [1.4K]

Answer:

b. $\Delta KE = 390 eV$

Explanation:

As we know that the electric field due to infinite line charge is given as

$E =\frac{\lambda}{2\pi \epsilon_0 r}$

here we can find potential difference between two points using the relation

$\Delta V = \int E.dr$

now we have

$\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr$

now we have

$\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})$

now plug in all values in it

$\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})$

$\Delta V = 216ln6 = 387 V$

now we know by energy conservation

$\Delta KE = q\Delta V$

$\Delta KE = (e)(387V) = 387 eV$

3 0

2 years ago

Help with physics projectile motion

BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0

3 years ago

A new ski area has opened that emphasizes the extreme nature of the skiing possible on its slopes. Suppose an ad intones "Free-f

Allushta [10]

Answer:

48.6°

Explanation:

The forward force, F equals the component of the weight along the slope.

So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.

So a = gsinθ

Since we are given that a = 75%g = 0.75g,

0.75g = gsinθ

sinθ = 0.75

θ = sin⁻¹(0.75)

= 48.6°

5 0

2 years ago

A control clinic offers a program that guarantees a weight loss of up to 0.46 kg in one week. Express the weight loss in a ratio

Mice21 [21]

Answer:

0.76 mg/s

Explanation:

0.46 kg/week × (1 week / 7 days) × (1 day / 24 hrs) × (1 hr / 3600 s) × (1000 g/kg) × (1000 mg/g) = 0.76 mg/s

4 0

2 years ago

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10

bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

$m_1$ = Mass of sphere = 10000 kg

$m_2$ = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

$\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s$

Angular momentum is given by

$L=I\omega$

Moment of inertia of the satellite is

$I_s=\frac{2}{5}m_1r^2$

Moment of antenna of the satellite is

$I_a=\frac{1}{3}m_2l^2$

The angular momentum of the system is

$L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s$

The angular momentum of the satellite is 605447.7066 kgm²/s

5 0

3 years ago