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2 years ago

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

Physics

1 answer:

scZoUnD [109]2 years ago

7 0

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

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Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p

Brums [2.3K]

Answer:

$P = 3.55 \times 10^5 Pa$

Explanation:

As we know that water from the fountain will raise to maximum height

$H = 26.0 m$

now by energy conservation we can say that initial speed of the water just after it moves out will be

$\frac{1}{2}mv^2 = mgH$

$v = \sqrt{2gH}$

$v = \sqrt{2(9.81)(26)}$

$v = 22.6 m/s$

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

$P = P_0 + \frac{1}{2}\rho v^2$

$P = 10^5 + \frac{1}{2}(1000)(22.6^2)$

$P = 3.55 \times 10^5 Pa$

6 0

2 years ago

Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc

Advocard [28]

Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax = 27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16 sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

= √27²+16² - (2(27)(16)(cos 60))

= 23.52 m

now we can use the sines law to find the angle

tan θ = $\frac{R_{y} }{R_{x} }$

= By - Ay/Bx -Ax

= (-8 - 13.5)/(-13.9 - (-23.4))

θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

= 24° east of south

4 0

2 years ago

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0

2 years ago

An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with

a_sh-v [17]

Explanation:

We assume that length of boat is L cm. And, we suppose to have a coordinate system relative to the bottom of the lake with:

Boat's position initially from x = 0 cm to x = L cm

Let person (S) is initially at x = L cm (right hand end of boat)

and, let boat's center of mass at x = C cm

Therefore, initially center of mass of overall (system's) is at $X_{1}$ where (taking moments about x = 0).

$(80 + 130)X_{1}$

= 130C + 80L ........... (i)

After S walks left to the other end of the boat, the boat moves right 80 cm.

Hence,

Boat's position is now from x = 80 cm to x = (L+ 80) cm

S's position is x = 80 (left end of boat)

Boat's center of mass is at (C + 80).

Therefore, overall (system's) final center of mass is at $X_{2}$ where (taking moments about x = 0).

$(80 + 130)X_{2}$

= $(130)(C + 80) + 80 \times 80$ ............. (ii)

As no external force is acted on the system, the system's center of mass has not moved, so $X_{1} = X_{2}$ .

This means the left hand sides of both equations (i) and (ii) are equal. So, we can equate the right hand sides giving: the equation as follows.

130C + 80L = $(130)(C+80) + 80 \times 80
$

80L = 16800 cm

L = 210 cm

= 2.1 m

Length of boat = 2.1 m

When the person is walking from one end to the other as there was no external force then center of mass of the person-boat system did not move.

6 0

2 years ago

The absence of any mechanical linkage between the throttle pedal and the throttle body requires the use of a _______ motor.

Rom4ik [11]

The answer is A.

hope it helps :)

8 0

2 years ago