2

A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rota

wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

Length of beam = 4.0 m

Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

Point at which person stands = 1.50 m from wall

Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

T×sin(53) = 350×1.5 + 200×2

T = 850/sin(53) = ‭1,064.315 N.

4 0

2 years ago

Which of the following is the smallest conceivable amount of time that could pass between a lunar eclipse and a solar eclipse .

pickupchik [31]

The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
because earth is making shade on moon

in order for solar eclipse to occur the line is now:
Sun-Moon-Earth
because moon is making a shade on earth (blocking sun = solar eclipse)

Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14

5 0

2 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance x that it covers after time t is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction θ such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at t ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$