A variable is defined as anything that

Physics

1 answer:

LenKa [72]2 years ago

3 0

Answer : Quantity or quality that varies like manipulating a variable to see what will happen to another variable

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In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea

Vikentia [17]

For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).

fringe = (delta t) / (λ/c)

We can find (delta t) with the equation:

delta t = [v^2(L1+L2)]/c^3

Derivation of this formula can be found in your physics text book. From here we find (delta t):

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.

4 0

3 years ago

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula

velikii [3]

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

$\rho$ = Density of fluid = $1.1\times 10^3\ kg/m^3$ (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = $\frac{4}{3}\pi r^3$

The weight of the bowling ball will balance the buouyant force

$W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg$

The mass of the bowling ball will be 6.1328 kg

Weight will be $6.1328\times 9.81=60.16284\ N$

5 0

2 years ago

When touched by a plastic straw, the metal sphere will do what?

bearhunter [10]

I believe that if you touch a metal sphere with a plastic straw, the straw would not have enough strength to push it. So in that case, the metal sphere will not move and will stay in one place.

Please rate a 5 star

5 0

2 years ago

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$