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Varvara68 [4.7K]

3 years ago

9

if an object is moving with constant speed then its distance time graph will be a straight line. a) along time axis b) along dis

tance axis c) parallel to time axis d)inclined to time axis.

1 answer:

Varvara68 [4.7K]3 years ago

5 0

Answer: I’m not sure but the time and distance will be increasing while they required the distance time so my answer is ((A))

Explanation:

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In a rectangular coordinate system a positive point charge q = 6.00 x 10^-9C is placed at the point x = +0.150 m, y = 0, and an

makvit [3.9K]

Answer:

a) 0N/C

b) 2660.7N/C

c) 543.7N/C

Explanation:

To find the electric field in all these cases you take into account the x and y component of the total Electric field for each point.

a)

for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:

$E_x=E_{1x}-E_{2x}\\\\E_x=k\frac{q}{r_1^2}-k\frac{q}{r_2^2}$ ( 1 )

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r1: distance to the first charge = 0.150m

r:2 distance to the second charge = 0.150m

Due to in this case the distance r1=r2: you obtain, by replacing in (1):

$E_x=0$

b)

for P(0.300 , 0 ) you also have only the x component of E, and the electric field generated by each charge are directed toward right:

$\vec{E}=E_x\hat{i}$

$E_x=k\frac{q}{r_1^2}+k\frac{q}{r_2^2}$

r1 = 0.300m+0.150m = 0.450m

r2 = 0.300m-0.150m = 0.150m

By replacing r1 and r2 in ( 1 ) you obtain:

$E_{x}=kq(\frac{1}{r_1^2}+\frac{1}{r_2^2})=(8.98*10^9Nm^2/C^2)(6.00*10^{-9})(\frac{1}{(0.450m)^2}+\frac{1}{(0.150m)^2})\\\\E_x=2660.7N/C\\\\\vec{E}=2660.7N/C\ \hat{i}$

c)

for P(0.150 , -0.40) you have both x and y components for E:

$\vec{E}=E_x\hat{i}+E_y\hat{j}\\\\E_x=k\frac{q}{r_1^2}cos\theta+0N/C\\\\E_y=-k\frac{q}{r_1^2}sin\theta-k\frac{q}{r_2^2}$

the second charge does not contribute for the x component of E.

To find r1 you use Pitagora's theorem:

$r_1=\sqrt{(0.150+0.150m)^2+(0.40m)^2}=0.500m$

r2 = 0.40m

the angle is obtain by using a simple trigonometric relation:

$tan\theta=\frac{0.40}{0.150}=2.66\\\\\theta=tan^{-1}(2.66)=69.44\°$

Then, by replacing the values of r1, r1, q, theta and k you obtain:

$E_x=(8.98*10^9Nm^2/C^2)\frac{(6.00*10^{-9}C)}{(0.500m)^2}cos69.44=75.68N/C\\\\E_y=-(8.98*10^9Nm^2/C^2)(6.00*10^{-9}C)(\frac{sin69.44}{(0.500m)^2}+\frac{1}{(0.40m)^2})\\\\E_y=538.42N/C\\\\\vec{E}=75.68N/C\hat{i}-538.42N/C\hat{j}\\\\|\vec{E}|=\sqrt{(E_x)^2+(E_y)^2}=543.7N/C\\\\\theta=tan^{-1}(\frac{538.42}{75.68})=278\°$

hence, the magnitude of E is 543N/C with an angle of 278° from the positive x axis.

4 0

3 years ago

A projectile is launched horizontally at a speed of 40 meters per second from a platform located a vertical distance h above the

KATRIN_1 [288]

Answer:1

Explanation:t=rad2h/g

4 0

3 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

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

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

3 years ago

A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the

Phantasy [73]

Answer:

3.4 x 10^-4 T

Explanation:

A = 1.5 x 10^-3 m^2

N = 50

R = 180 ohm

q = 9.3 x 106-5 c

Let B be the magnetic field.

Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .

Let e be the induced emf and i be the induced current

e = rate of change of magnetic flux

e = dФ / dt

i / R = B x A / t

i x t / ( A x R) = B

B = q / ( A x R)

B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T

6 0

4 years ago

Electron (-) has a charge of 1.6 x 10-19 C. Proton (+) has a charge of 1.6 x 10-19 C. In a hydrogen atom the distance between a

gladu [14]

Answer:

c because you can lift a 2 pound dumbell

7 0

4 years ago