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Montano1993 [528]
2 years ago
7

The aircraft link is made from an a992 steel rod. determine the smallest diameter of the rod, to the nearest 1 /16 in., that wil

l support the load of 4 kip without buckling. the ends are pin connected.
Physics
1 answer:
Bas_tet [7]2 years ago
3 0

Answer:

hjhjhjhj

Explanation:

jhjhjhjhjh

You might be interested in
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x
mezya [45]

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

3 0
2 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
frez [133]

Answer:

<em>1.49 x </em>10^{11}<em></em>

<em></em>

Explanation:

Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>

Mathematically, this can be stated as

T^{2} ∝ R^{3}

<em>to remove the proportionality sign we introduce a constant</em>

T^{2} = kR^{3}

k = \frac{T^{2} }{R^{3} }

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x 10^{11}

for mars, constant k will be

k = \frac{687^{2} }{(2.279*10^{11}) ^{3} } = 3.987 x 10^{-29}

For Earth, orbital period T is 365 days, therefore

365^{2} = 3.987 x 10^{-29} x R^{3}

R^{3} = 3.34 x 10^{33}

R =<em> 1.49 x </em>10^{11}<em></em>

8 0
2 years ago
Read 2 more answers
1. The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force th
Ludmilka [50]

Answer:

1. Earth pulls the sun away from itself with an equal force.

6 0
2 years ago
Read 2 more answers
What affects the way a projectile performs when it is shot from a firearm?
adoni [48]

Answer:

In no particular order, they are:

Propellant: The amount of propellant and the type of propellant will determine the ultimate range of the projectile.  

Muzzle Length: In accordance with propellant, determines the muzzle velocity of the projectile. A general rule of thumb is, the longer the muzzle, the higher the muzzle velocity.

Projectile Type: As in it's shape. A baseball is not going to fly as far as an artillery shell, even if thrown at the same speeds, due to the drag it creates.

Atmospheric Pressure: Affects how dense the air is, determines how much drag the projectile will have to fly through, affecting it's range.

Temperature: Same as atmospheric pressure.

Wind: Depending on the speed and direction, can result in the projectile arriving in places it has no business being in.

Humidity: Same as atmospheric pressure, but also if a projectile has to travel through a particularly humid area when the conditions are just right and if the projectile is in the air long enough, it can accumulate ice, affecting it's shape, increasing drag and increasing it's weight.

Spin of the Earth, or The Coriolis Effect: Shows it's effects most at the West-East, and vice versa, directions. Can be negligible in short distances, such as when a pistol is fired or a ball is thrown. Will become more noticeable as the range increases and will need corrections to aiming such as when firing a sniper rifle at long range or artillery fire. Projectiles fired towards the East will end up higher than the aim point, and those that're fired West will end up lower. In long enough ranges it can cause the projectile to miss a whole city.

Spin of the Projectile: Almost all long range artillery and firearms will induce a spin on the projectile through rifling in the gun barrel. Rifling and the rate of spin will depend on the particular gun and it's design/purpose. Spinning a projectile will make it more stable in flight, ergo making it more accurate. But that spin comes from redirecting some of the kinetic energy during propulsion and will cause the projectile to have less range compared to an identical smoothbore gun with the same propellant.

Gravity: Not a major consideration for most applications on Earth truth be told, but is crucial when said projectile is designed to go to space, like rockets.

These are what I could remember with sleepy eyes, having just woken up. It should be comprehensive enough, but if I remember one I missed, I'll be sure to add it here.

6 0
2 years ago
Read 2 more answers
A girl at an airport rolls a ball north on a moving walking that moves east. If the ball’s speed with respect to the walkway is
sweet-ann [11.9K]

We have to add two vectors.

Vector #1: 0.15 m/s north

Vector #2: 1.50 m/s east

Their sum:

Magnitude: √(0.15² + 1.50²)

Magnitude = √(0.0225+2.25)

Magnitude = √2.2725

Magnitude = <em>1.5075 m/s</em>

Direction = arctan(0.15/1.50) north of east

Direction = <em>5.71° north of east</em>

4 0
2 years ago
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