PHYSICS COLLEGE

6. A barber raises his customer's chair by applying a force of 150 N to thehydraulic piston of area 0.01 m2. If the chair is attached to a piston with an
area of 0.1 m², how much force is needed to raise the customer?
STEP 1: List the known
and unknown values F =
A=
A,
STEP 2: Write the
correct equation
STEP 3: Insert the
known values into the
equation to solve for
the unknown value

Answers

Answer 1

Answer:

Answer:

15N

Explanation:

F¹=150N

A=0.01m2²

F2=?

A2=0.1m²

P=F/A

F1/A2=F2/A1

150/0.1=F2/0.01

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are
(b) the total work done on it,
(c) the work done by the gravitational force on the crate, and
(d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Answers

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = (d )/(L)$

=> $\theta = Sin^(-1) (d)/(L)$ =

Now substituting thje values

=> $\theta = Sin^(-1) (4)/(12)$ =

=> $\theta = Sin^(-1) (1)/(3)$

=> $\theta = Sin^(-1)(0.333)$

=> $\theta = 19.5^(\circ)$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = (mg)/(cos\theta)$

=> $T = (230 * 9.8 )/(cos(19.5))$

=> $T = (2254 )/(cos(19.5))$

=> $T = (2254 )/(0.9426)$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 * 9.8* 12 ( 1 - cos(19.5) )$

= $-230 * 9.8* 12 ( 1 - 0.9426) )$

= $-230 * 9.8* 12 (0.0574)$

= $-230 * 9.8* 0.6888$

= $-230 * 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. If cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.

Answers

Find the given attachments

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

Answers

Answer:

$276.74* 10^8Mg/m^3$

31.29 m/sec

Explanation:

We have given density of substance $0.14lb/in^3$

We have convert this into $Mg/m^3$

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram $=63.49* 10^3\ Mg$

We know that $1 in^3=1.6387* 10^(-5)m^3$

So $0.14in^3=0.14* 1.6387* 10^(-5)=0.2294* 10^(-5)m^3$

So $0.14lb/in^3$ =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr $=(70* 1609.34meter)/(3600sec)=31.29m/sec$

A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?

Answers

Answer:

Energy, $E=4.002* 10^(-19)\ J$

Explanation:

It is given that,

Wavelength of the photon, $\lambda=500\ nm=5* 10^(-7)\ m$

We need to find the photon representing the particle interpretation of this light. it is given by :

$E=(hc)/(\lambda)$

$E=(6.67* 10^(-34)* 3* 10^8)/(5* 10^(-7))$

$E=4.002* 10^(-19)\ J$

So, the energy of the photon is $4.002* 10^(-19)\ J$ . Hence, this is the required solution.

A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye

Answers

Answer:

The value is $p = - 2.63 \ Diopters$

Explanation:

From the question we are told that

The value of the far point is $a = 40 \ cm = 0.4 \ m$

The distance of the lens to the eye is $b = 2 \ cm = 0.02$

Generally

$1 Diopter = > 1 m^(-1)$

Generally the power spectacle lens needed is mathematically represented as

$p = (1)/(d_o ) + (1)/(d_i)$

Here $d_o$ is the object distance which for a near sighted person is $d_o = \infty$

And $d_i$ is the image distance which is evaluated as

$d_i = b - a$

=> $d_i = 0.02 - 0.4$

=> $d_i = -0.38 \ m$

$p = (1)/(\infty ) + (1)/(-0.38)$

=> $p = 0 - 2.63$

=> $p = - 2.63 \ Diopters$

Why is there so much more carbon dioxide in the atmosphere of Venus than in that of Earth? Why so much more carbon dioxide than on Mars?

Answers

Explanation:

The reason for the more concentration of carbon dioxide in the atmosphere of Venus than in the Earth -

On the Earth , most amount of the carbon dioxide is in the ocean water and in sea sediments .

Considering Venus , in the planet Venus , there is no Ocean water , hence , carbon dioxide can not get dissolved into the water , hence , it is found in the atmosphere .

So , the escape velocity for carbon dioxide on Mars is smaller than Venus .

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