If a male runner in the fourth starting position ran the 400 m race in 44.40 s, how would you

(35 points!!) A block and tackle is used to lift a load weighing 600 lb. The operator lifts the load distance of 1.25 ft by pull

Drupady [299]

Answer:

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

Explanation:

Given:

Weight of the load, $W=600$ lb.

Distance moved by the load, $D_{load}=1.25$ ft.

Force applied on the rope, $F = 200$ lb

Distance moved by the force on the rope, $D{in}=4$ ft.

Work done by a force causing a displacement in the direction of force is given as:

$\textrm{Work}=\textrm{Force}\times \textrm{Displacement}$

A.

Here, Input Work is given by the input force and the displacement caused by the input force. So,

$W_{in}=F\times D_{in}\\W_{in}=200\times 4=800\textrm{ ft.lb}$

Therefore, the input work is 800 lb.ft

B.

Output Work is given by the output force and the displacement caused by the output force. So,

$W_{out}=F_{load}\times D_{load}\\W_{out}=600\times 1.25=750\textrm{ ft.lb}$

Therefore, the output work is 750 ft.lb

C.

Efficiency is given as the ratio of Output Work to Input Work expressed as percentage. So,

Efficiency = $\frac{W_{out}}{W_{in}}\times 100=\frac{750}{800}\times 100=93.75\%$

Therefore, efficiency of block and tackle is 93.75 %.

6 0

3 years ago

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Which statement describes the formation of the Milky Way galaxy?

Oliga [24]

Answer:

A disk formed of a long trail of stars coiled into a spiral.

Explanation:

5 0

3 years ago

How are Piaget and Erikson's theories different?

SOVA2 [1]

D) all of the above is the answer

8 0

3 years ago

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

$J=\frac{E}{\rho}$ (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

$J=\frac{I}{A}$ (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

$\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}$

Next, you replace for all variables:

$I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A$

hence, the current in the wire is 4.75A

4 0

3 years ago

Define volume charge density <img src="https://tex.z-dn.net/?f=%20%28%20%5C%3A%20%5Crho%20%5C%3A%20%29%20" id="TexFormula1" titl

azamat

Answer:

Volume charge density: When the charge density is throughout the volume of a body, then the charge per unit volume is called volume charge density and it is represented by ρ.

∴ρ=

$\frac{v}{p}$

Hence the unit of ρ is Cm

−3

Example: If a charge q is uniformly distributed in the whole volume of a

sphere of radius R, then p =

$\frac{q}{ \frac{4}{3} } \pi {r}^{3}$ =3q/4πR-3

3 0

2 years ago

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