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3 years ago

Compared to a plane gravitational pull on the ground to 7 miles in sky is what

1 answer:

Citrus2011 [14]3 years ago

4 0

When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.

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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

Andy has two strong magnetic marbles of equal size. What will happen when he puts the two marbles about 2 centimeters apart?

VikaD [51]

I believe it would be C

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The top priority in a lab setting is

SpyIntel [72]

The answer is a.safety

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3 years ago

Just think about this.

diamong [38]

This ain’t the place, bud. If you have a QUESTION, then you can post it here.

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A small branch is wedged under a 200 kg rock and rests on a smaller object. The small object that creates a pivot point is calle

KiRa [710]

Answer:

F = 326.7 N

Explanation:

given data

mass m = 200 kg

distance d = 2 m

length L = 12 m

solution

we know force exerted by the weight of the rock that is

W = m × g ..............1

W = 200 × 9.8

W = 1960 N

and

equilibrium the sum of the moment about that is

∑Mf = F(cos∅) L - W (cos∅) d = 0

here ∅ is very small so cos∅ L = L and cos∅ d = D

so F × L - W × d = 0 .................2

put here value

F × 12 - 1960 × 2 = 0

solve it we get

F = 326.7 N

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3 years ago

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