The blank represent the different forms of electromagnetic radiation

(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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A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle

Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

$\tau=NIABsin\theta$ (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

$B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T$

3 0

3 years ago

A student sees a newspaper ad for an apartment that has 2030 square feet (ft2) of floor space. How many square meters of area ar

Vilka [71]

Square feet or square meters?

8 0

3 years ago

Gamma rays are waves of energy that have no charge. <br> True<br> False

Andre45 [30]

Gamma rays are waves of energy that have no charge. TRUE.

3 0

3 years ago

Read 2 more answers

1. A jet lands on an aircraft carrier at a speed of 140 mi/h [ 1.00 mile = 1.61 km]

Iteru [2.4K]

Answer:

Acceleration, a = -31.29 m/s²

Explanation:

It is given that, Initial speed of the aircraft, u = 140 mi/h =

62.58 m/s

Finally, it stops, v = 0

Time taken, t = 2 s

Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :

a = t

0-62.58 2 a =

a = -31.29 m/s²

So, the acceleration of the aircraft is 31.29 m/s²

and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)

5 0

3 years ago