A string is stretched by two equal and opposite forces F newton each then the tension in sting is?

Physics

1 answer:

Arlecino [84]3 years ago

3 0

Answer: The tension in the string is zero

Explanation:

You might be interested in

How do I calculate initial speed by using the slope of a graph? The graph shows the relationship between S {(sin2ϴ)/g} and the a

Orlov [11]

The formula for slope is y=mx+b

3 0

3 years ago

A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above

Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + $\frac{1}{2}at^2$ and plugging in from the y-dimension info:

$-10.0=0t+\frac{1}{2}(-9.8)t^2$ which simplifies to

$-10.0=-4.9t^2$ so

$t=\sqrt{\frac{-10.0}{-4.9} }$ which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in the x-stuff:

Δx = $7.25(1.4)+\frac{1}{2}(0)(1.4)^2$ Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation: