Answer:
F = - K x
a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2
b) ω = (K/m)^1/2 angular frequency of SHM
ω = (133 / 1.3)^1/2 = 10.1 / sec
f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec
P = 1/f = .0157 sec
Da ne ajuta pentru a putea citi corect lungimea ... coronită???
Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:


Now we find the centripetal acceleration which is given by

where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find

we also have a tangential acceleration, which is given by

where
t = 17.0 s
Substituting values,

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

Answer:
a . 0.35cm
b. 11.33cm
Explanation:
a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

Hence, for currents in same direction, the point is 0.35cm
b. Given both currents flow in opposite directions, the null point lies on the other side.
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:
Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

Hence, if currents are in opposite directions the point on x-axis is 11.33cm
The answer is: [C]: "4" .
___________________________________________________
Note: To balance this equation, the coefficient, "4", should be placed in front of the PCl₃ ; and the coefficient, "6", should be placed in front of the Cl₂ .
________________________________________________________
The balanced equation is:
__________________________________________________
P₄ (s) + 6 Cl₂ (g) <span>→ 4 </span>PCl₃ (l) .
______________________________________________________