Ask question

Ask question

All categories

4 days ago

8

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

hat is the kinetic energy of this charge at x = 4m?

1 answer:

ValentinkaMS [2.4K]4 days ago

8 0

The force due to electricity on the charge is calculated by multiplying the charge by the intensity of the electric field:
$F=qE$
in our scenario, where $q=3 \mu C= 3 \cdot 10^{-6} C$ and $E=2 kN/C=2000 N/C$ , resulting in the force of
$F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N$

Initially, the kinetic energy of the particle is at zero (as it remains stationary), which means its final kinetic energy is equal to the work performed by the electric force over a distance of x=4 m:
$K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J$

You might be interested in

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

Ostrovityanka [2204]

Response:

y= 240/901 cos 2t+ 8/901 sin 2t

Clarification:

To determine mass m=weight/g

m=8/32=0.25

To calculate the spring constant

Kx=mg (with c=6 inches and mg=8 pounds)

K(0.5)=8 (6 inches converts to 0.5 feet)

K=16 lb/ft

The governing equation for the spring-mass system is

my''+Cy'+Ky=F

Inserting the known values yields

0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given C=0.25 lb.s/ft)

Assuming the steady state equation for y is

y=A cos 2t+ B sin 2t

To determine constants A and B, we must equate this with equation 1.

Next, we find y' and y" by differentiating with respect to t.

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now, substitute the values of y", y' and y into equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

By comparing coefficients on both sides

30 A+ B=8

A-30 B=0

From this, we find

A=240/901 and B=8/901

Thus, the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0

10 days ago

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

Softa [2029]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; $\sum fy = ma_y$

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

3 0

27 days ago

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

serg [2593]

Answer:

$v = 3369.2 m/s$

Explanation:

The beacon is rotating at an angular speed of

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

We know that

$v = r \omega$

At this point we have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

So we can conclude with

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

19 days ago

Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st

Ostrovityanka [2204]

Answer:

d) v1 = v2 = v3

Explanation:

This can be determined through the principle of energy conservation. We assess the total mechanical energy E=K+U (the sum of kinetic energy and gravitational potential energy) at both the initial and final positions, ensuring they remain constant.

<pInitially, for the three spheres, we have:

$E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh$

Finally, for the three spheres, we see:

$E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}$

<pGiven that $E_i=E_f$ , and since $E_i$ remains identical for all spheres, it follows that $E_f$ is identical for all spheres, indicating that $v_f$ , the final velocity, is equal for each ball.

4 0

27 days ago

A friend climbs an apple tree and drops a 0.22-kg apple from rest to you, standing 3.5 m below. When you catch the apple, you br

Keith_Richards [2256]

(A) velocity = 2.8 m/s (B) Average force = 1.9536 Newtons.

3 0

6 days ago