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3 years ago

We dont got the force so i got no clue what to do

Physics

1 answer:

konstantin123 [22]3 years ago

3 0

Answer:

work = 1275.3 J

Explanation:

work = (force)(distance)cosø ------- force = ma

=(mass*acceleration)(distance)cosø

=(20*9.81)(6.5)cos0

=1275.3J

nite that the angle of cosine is the difference between the angle of force and the distance. in this case, the force and the distance are in the same direction. :)

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When air cools, it holds _____ water than when it is warmer.

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Which is evidence that a convergent boundary once existed

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Volcanic islands, Mountain ranges

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2 years ago

We are given the description of an enclosure that has fencing on 3 sides and can use 3,056 yards of fencing. We are to find the

shusha [124]

Answer:

The sum of the lengths of the sides is 2292 yards and the sum of the lengths of the triangle is 3056 yards

Explanation:

Since y represents the length of fence that is opposite (parallel) to the river and x represent the length of fence perpendicular to the river.

Therefore since we can use 3,056 yards of fencing

Side perpendicular to the river = x and,

Side opposite to the river = y = 3056 - 2x

The area of the rectangle formed (A) = Perpendicular side × Parallel side

∴ A = x(3056 - 2x) = 3056x - 2x²

A = 3056x - 2x²

To maximize the area, A' (dA/dx) = 0

∴ A' = 3056 - 4x = 0

3056 - 4x = 0

4x = 3056

x = 764 yards

y = 3056 - 2x = 3056 - 2(764) = 1528 yards.

Side perpendicular to the river = 764 yards and,

Side opposite to the river = 1528 yards

The sum of the lengths of the sides = 764 + 1528 = 2292 yard and the sum of the lengths of the triangle = 764 + 764 + 1528 = 3056 yards

5 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago