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katen-ka-za [31]
3 years ago
13

1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic​

Physics
1 answer:
Korvikt [17]3 years ago
8 0

51.448 g is the required answer!

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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

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Coronary artery disease is a disease of the heart where the arteries and blood vessels become clogged with fat deposits called _
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A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

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a ball is thrown with an initial velocity of 80 feet per second, at an angle of 40 degrees with the horizontal. Find the vertica
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Well velocity is a vector quanity and we can brake vectors in to verticle and horizantal componontes threw these two formulas Vcostheta=Vx and Vsintheta=Vy
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