All categories

3 years ago

HURRY

Physics

1 answer:

exis [7]3 years ago

5 0

The answer is the last choice.

Its electrical potential energy stays the same because it has the same electric potential. The reason why is that moving the charge towards X does not change the distance of the negative charge between the plates. The Electrical potential energy of a particle is the result energy by virtue of its position from the electrical fields produce by the plates both positive and negative. Since the charge is still equidistant to each other (assuming based from the diagram) no change in terms of electrical energy consumption or work was done.

You might be interested in

The main water line enters a house on the first floor. The line has a gauge pressure of 1.94 x 105 Pa. (a) A faucet on the secon

Ad libitum [116K]

Answer

given,

gauge pressure = 1.94 x 10⁵ Pa

Pressure due to 4.90 m column of water

= ρ g h

= (4.90) x (1000) x (9.8) Pa

= 48020 Pa

Gauge pressure of second floor faucet

= 1.94 x 10⁵Pa - 48020 Pa

P_g= 145980 Pa

( b )

Let h = height of faucet from which no water can flow even if open

P = ρ g h

1.94 x 10⁵ = h x(1000) x (9.8)

h = 19.79 m

4 0

3 years ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x .........1

so force = mg = 0.35 (9.8) = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × $\sqrt{\frac{m}{k} }$ ................2

put here value

period of oscillations = 2π × $\sqrt{\frac{0.35}{28.58} }$

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0

3 years ago

Gawaingnpang nkabuhayan,hamon at oportinidad

Savatey [412]

gawaingnpang nkabuhayan, hamon at oportinidad

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

What type of clouds are associated with low pressure?

Naddik [55]

Cumulus and cumulonimbus<span />

5 0

3 years ago

***HURRY***

HURRY