Please help me please

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

trapecia [35]

Initial speed of Gazelle is along x direction and its value will be

$v_x = 5.6 m/s$

also its initial height is given as

$y = 2.5 m$

Part a)

now from kinematics along Y direction

$\Delta y = v_y t + \frac{1}{2} at^2$

as we know that

$\Delta y = 0$

$v_y = 0$

$a = 9.8 m/s^2$

$2.5 = 0 + \frac{1}{2} (9.8) t^2$

$t = 0.714 s$

Part b)

distance moved horizontally

$\Delta x = v_x t$

as we know that

$v_x = 5.6 m/s$

now we will have

$v_x = 5.6 (0.714) = 4m$

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

$v_{fy} = v_y + at$

$v_{fy} = 0 + (9.8)(0.714) = 7 m/s$

$v_x = 5.6 m/s$

so net speed will be

$v^2 = v_x^2 + v_y^2$

$v^2 = 7^2 + 5.6^2$

$v = 8.96 m/s$

7 0

4 years ago

The loudness of a sound is related to the _____ of the sound wave. wavelength frequency velocity amplitude

Vesna [10]

The answer is wavelength

6 0

4 years ago

Read 2 more answers

A light-year measures the _______ that light travels in 1 year.

o-na [289]

<span>A light-year measures the distance that light travels in 1 year.

Answer : B ) Distance

-Hope this helps.</span>

4 0

3 years ago

Read 2 more answers

a water line starts the service with an altitude of 1200m over the sea level, what is the velocity of the water above 1050 m ove

Ilia_Sergeevich [38]

Answer:

Velocity = 94.85m/s

Explanation:

<u>Given the following data ;</u>

Height = 1200m

Vertical distance = 1050m

To find the time, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting into the equation, we have;

$1200 = 0(t) + \frac {1}{2}*9.8*t^{2}$

$1200 = 0 + 4.9*t^{2}$

$1200 = 4.9*t^{2}$

$t^{2} = \frac {1200}{4.9}$

$t = \sqrt{122.45}$

t = 11.07 secs

To find the velocity;

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

Substituting into the above equation;

$Velocity = \frac{1050}{11.07}$

Velocity = 94.85m/s

Therefore, the velocity of the water above 1050 m over the sea level is 94.85m/s.

7 0

3 years ago

Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the

Svetlanka [38]

Answer:

Kindly check the explanation section.

Explanation:

For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.

From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.

Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.

Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).

If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.

The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500

The fracture strength = .75 × Ah × Fhb = 309 kips.

The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.

4 0

4 years ago