Answer: g = 10.0 m/s/s
Explanation:
For a simple pendulum, provided that the angle between the lowest and highest point of his trajectory be small, the oscillation period is given by the following expression:
T = 2π √L/g , where L = pendulum length, g= accelleration of gravity.
We can also define the period, as the time needed to complete a full swing, so from the measured values, we can conclude the following :
T = 140 sec/ 101 cycles = 1.39 sec
Equating both definitions for T, we can solve for g, as follows:
g = 4 π² L / T² = 4π². 0.49 m / (1.39)² = 10.0 m/s/s
Resultant force is basically the force left after everything is added.
if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:
180N - 84N =<u> 96N</u> (you can determine whether it's positive or negative based on the direction of the vector)
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer: 2.49×10^-3 N/m
Explanation: The force per unit length that two wires exerts on each other is defined by the formula below
F/L = (u×i1×i2) / (2πr)
Where F/L = force per meter
u = permeability of free space = 1.256×10^-6 mkg/s^2A^2
i1 = current on first wire = 57A
i2 = current on second wire = 57 A
r = distance between both wires = 26cm = 0.26m
By substituting the parameters, we have that
Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26
= 4080.744×10^-6/ 1.634
= 4.080×10^-3 / 1.634
= 2.49×10^-3 N/m
Answer: Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Another bright star, Regulus, has a parallax of 0.042 arcseconds. Then, the distance in parsecs will be,23.46.
Explanation: To find the answer, we have to know more about the relation between the distance in parsecs and the parallax.
<h3>What is the relation between the distance in parsecs and the parallax?</h3>
- Let's consider a star in the sky, is d parsec distance from the earth, and which has some parallax of P amount.
- Then, the equation connecting parallax and the distance in parsec can be written as,
<h3>How to solve the problem?</h3>
- Thus, we can find the distance in parsecs as,
Thus, we can conclude that, the distance in parsecs will be, 23.46.
Learn more about the relation connecting distance in parsecs and the parallax here: brainly.com/question/28044776
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