__________ refers to an object positioned in the back of something or somebody ?

PLZ HELP ME ITS MY LAST QUESTION I GOT LIKE 6 MIN LEFT

spin [16.1K]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height = 8.18m

Unknown:

Speed of the diver before hitting the water = ?

Solution:

The speed of the diver before hitting the water is the final velocity.

To solve this problem, we use the expression below;

v² = u² + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

Now insert the parameters and solve;

v² = 0² + 2 x 9.8 x 8.18

v² = 160.328

v = 12.7m/s

3 0

3 years ago

Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel

andrew-mc [135]

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

Por la definición de trabajo sabemos que el montañero debió contrarrestar trabajo causado por la gravedad terrestre. Si asumimos que el cambio de la altura es muy pequeño en comparación con el radio del planeta (6371 kilómetros vs. 0,5 kilómetros), entonces podemos considerar que la aceleración gravitacional es constante y la ecuación de trabajo ( $\Delta W$ ), medido en joules, que reducida a:

$\Delta W = m\cdot g\cdot \Delta z$ (1)

Donde:

$m$ - Masa del montañero, medido en kilogramos.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$\Delta z$ - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que $m = 65\,kg$ , $g = 9,807\,\frac{m}{s^{2}}$ y $\Delta z = 500\,m$ , entonces el trabajo realizado por el montañero para subir es:

$\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)$

$\Delta W = 318727,5\,J$

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

7 0

3 years ago

3 simple machines that make up a can opener

Yanka [14]

Hello There

Answer: The wheel and axle, the lever, and the wedge

Reason: The two long arms that clamp onto the can are levers. The handle which is used to rotate the can is the wheel and axle<span>. Finally the blade is the wedge

I hope I helped
-Chris</span>

6 0

3 years ago

Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?

LUCKY_DIMON [66]

Answer:NOOPE

Explanation:IM THE MYSTERY MAN WHOOOOSHHHHHHHHHHHHHHH ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

6 0

3 years ago

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago