We need first to use the formula F=m(a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be
<span>m=2kg(both blocks)+500g(both ropes) → m=2.5kg </span>
<span>a=3.00m/s^2 </span>
<span>g=9.8m/s^2 </span>
<span>F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N
To calculate the tension at the top of rope 1 you need to use the formula </span>T=m(a+g) so it will be <span>T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N
</span>We can now calculate the tension at the bottom of rope 1 using the formula: <span>T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N
</span>Now to find the tension at the top of rope 2 we do it like this:
<span>T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2</span>
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
Answer:
It slowly decreases and the friction acting on it slowing it down becomes the bigger net force, if that makes sense :)
Explanation:
The dimension of K is M/ T^2
according to the question T=2π square root ofm/k here 2 pi is constant so
T= root of m /k and root of k = root of m/ T now by squaring on both the sides we get the answer k= M/ T^2
complete question :
A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T=√2πm/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?
To learn more about dimension:
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Answer:
Find the ideas below.
Explanation:
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