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2 years ago

a crane does 9,500 J of work to lift a crate straight up using a force of 125 N. how high does the crane lift the crate?

Physics

1 answer:

Lina20 [59]2 years ago

7 0

Answer:

<em>The crane lifts the crate up to 76 m high</em>

Explanation:

<u>Work Done by a Force</u>

The work done by a force of magnitude F that displaces an object by a distance y is given by

$W=F.y$

We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.

The above equation can be solved to know the value of y in terms of the work and the force:

$\displaystyle y=\frac{W}{F}$

Plugging in the given values

$\displaystyle y=\frac{9,500}{125}$

$y=76\ m$

The crane lifts the crate up to 76 m high

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Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below

MatroZZZ [7]

The effective height of the water for Smith's house will be 24.61m.

<h3>How to calculate the height?</h3>

Based on the information given, the volume of the water in sphere will be:

= 4/3πr³ = (5.80 × 10^5)/1000

= 4.18r³ = 580

r³ = 138.7

r = 5.18m

The effective height of the water will be:

= 18.0 + 2(5.18)

= 28.36

The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

= 18.0 + 2(5.18) - 3.75

= 24.61m

The gauge pressure at Faucet of Jones house will be:

= 1000 × 9.8 × 24.61

= 241.2kPa

Learn more about height on:

brainly.com/question/983412

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8 0

1 year ago

A veterinarian thinks that a dog has swallowed a key ring. Which types of electromagnetic waves is the doctor most likely

poizon [28]

Answer:

X rays

Explanation:

4 0

3 years ago

Read 2 more answers

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

2 years ago

Can anyone tell me how to read a micrometer screw gauge I want very clear instructions.

Natalka [10]

Explanation:

Things you need to know:

Accuracy refers to the maximum error encountered when a particular observation is made.

Error in measurement is normally one-half the magnitude of the smallest scale reading.

Because one has to align one end of the rule or device to the starting point of the measurement, the appropriate error is thus twice that of the smallest scale reading.

Error is usually expressed in at most 1 or 2 significant figures.

Tape

Equipment: It is made up of a long flexible tape and can measure objects or places up to 10 – 50 m in length. It has markings similar to that of the rigid rule. The smallest marking could be as small as 0.1 cm or could be as large as 0.5 cm or even 1 cm.

How to use: The zero-mark of the measuring tape is first aligned flat to one end of the object and the tape is stretched taut to the other end, the reading is taken where the other end of the object meets the tape.

Ruler

Equipment: It is made up of a long rigid piece of wood or steel and can measure objects up to 100 cm in length. The smallest marking is usually 0.1 cm.

How to use: The zero-end of the rule is first aligned flat with one end of the object and the reading is taken where the other end of the object meets the rule.

Vernier Caliper

Equipment: It is made up of a main scale and a vernier scale and can usually measure objects up to 15 cm in length. The smallest marking is usually 0.1 cm on the main scale.

It has:

a pair of external jaws to measure external diameters

a pair of internal jaws to measure internal diameters

a long rod to measure depths

How to use: The jaws are first closed to find any zero errors. The jaws are then opened to fit the object firmly and the reading is then taken.

Micrometer Screw Gauge

Equipment: It is made up of a main scale and a thimble scale and can measure objects up to 5 cm in length. The smallest marking is usually 1 mm on the main scale (sleeve) and 0.01 mm on the thimble scale (thimble). The thimble has a total of 50 markings representing 0.50 mm.

It has:

an anvil and a spindle to hold the object

a ratchet on the thimble for accurate tightening (prevent over-tightening)

How to use: The spindle is first closed on the anvil to find any zero errors ( use the ratchet for careful tightening). The spindle is then opened to fit the object firmly (use the ratchet for careful tightening) and the reading is then taken.

5 0

3 years ago

Bart stole a watermelon and ran 5,000 feet from the cops and they chase lasted 0.1 hours how fast was Bart running in miles per

liraira [26]

I am not as sure but I think it is 9.469 miles

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2 years ago