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Colt1911 [192]
3 years ago
9

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

Physics
1 answer:
scoundrel [369]3 years ago
7 0

Answer:

The answer is "5 \ \frac{m}{s} \ west"

Explanation:

\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to  \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\

velocity of car | respect to car :

\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\

          =\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west

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100m / 97.2m/s = 1.0288 seconds
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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

5 0
3 years ago
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

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3 years ago
A typical adult human has a mass of about 70 kg.
Misha Larkins [42]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2 
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b. </span><span>FE GmEmr2
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3 years ago
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