I load a 0.4 kg marble into a slingshot and shoot it directly upward by applying a 36.2 N force

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

8 0

3 years ago

Do mirrors reflect light

777dan777 [17]

Answer:

Yes, Mirror are a surface that reflects light more perfectly than ordinary objects.

Explanation:

8 0

3 years ago

if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for

bagirrra123 [75]

Gravitational force between two masses is given by formula

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = 55 kg$

$m_2 = 1234 kg$

$r = 3 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

now from the above equation we will have

$F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}$

$F = 5.03 \times 10^{-7}N$

so above is the gravitational force between car and the person

5 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

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