Ask question

Ask question

All categories

2 years ago

6

A simple wheel and axle is used to lift a bucket of water out of a well. the radii of the wheel and axle are 20cm and 4cm, respe

ctively. determine a) the theoreical effort required to lift a load of 30 N assuming no energy loss b) the efficiency if the actual effort required is 10 N

2 answers:

Veronika [31]2 years ago

5 0

Answer: i think it is A

Explanation: hope it helped

ddd [48]2 years ago

5 0

Answer:

A

Explanation:

need and explanation or no?

You might be interested in

If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply

Flauer [41]

Answer:

$T=8.33A$

Explanation:

From the question we are told that:

Number of battery $n=10$

Voltage source $E=12V$

Lamp Power $P=10W$

Generally the equation for Resistance is mathematically given by

$R=\frac{V^2}{P}$

$R=\frac{12^2}{10}$

$R=14.4ohms$

Therefore

$R_{eq}=\frac{14.4}{10}$

$R_{eq}=1.44$

Generally the equation for Current is mathematically given by

$T=\ffrac{V}{Req}$

$T=\frac{12}{1.44}$

$T=8.33A$

6 0

2 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

2 years ago

How would the force of a test charge change if the electric field is doubled?

Stolb23 [73]

<span>The force would double.</span>

7 0

3 years ago

Read 2 more answers

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$

$kg\dfrac{m^2}{s^2}=Joule$

$P=\dfrac{kgm^2}{s^3}$

So, the answer is D. the proper replacement unit for one joule per second

8 0

3 years ago

A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?

Montano1993 [528]

Net Force = (mass) x (acceleration) (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

6 0

2 years ago