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a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic system?

Answers

Answer 1

Answer:

Answer:

The answer is 7 Pa

Explanation:

The pressure transmitted in the hydraulic system can be found by using the formula

$p = (f)/(a) \n$

f is the force

a is the area

From the question we have

$p = (35)/(5) \n$

We have the final answer as

7 Pa

Hope this helps you

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A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?

Answers

(a) Length of the wire is 3.162 m

(b)Power delivered to the coil is 339.7 W

Electrical Power:

The electrical power is given by

P = V² / R

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 Ω

Now the resistivity of a wire is given by

ρ= RA/L

here ρ = 1.50×10⁻⁶ Ωm

so after rearranging we get:

L = RA / ρ

Now, the radius of wirer = 0.5 / 2 mm = 0.25 mm = 2.5×10⁻⁴ m

So the cross sectional area can be calculated as follows

$A = \pi r^2\n\nA = \pi * (2.5*10^(-4))^2\n\nA = 1.96*10^(-7) m^2$

hence,

$L = (24.2 *1.96*10^(-7) / 1.50*10^(-6)) \n\nL = 3.162\; m$

(b)The dependency of resistance with temperature is as follows:

R = R₀[1 + αΔT]

α = $4*10^(-4)^\;oC^(-1)$ for Nichrome

$R' = R [1 + \alpha (1200 - 20) ]\n\nR' = R[1 + \alpha (1180) ]\n\nR' = 24.2[ 1 + 4*10^(-4) * 1180 ]\n\nR' = 24.2[1 + 0.472]\n\nR' = 24.2 * 1.472\n\nR' = 35.62 \;\Omega$

So the power generated is :

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Learn more about electrical power:

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Answer:

a) 3.162 m

b) 339.7 W

Explanation:

Assume ρ = 1.50*10^-6 Ωm, and

α = 4.000 10-4(°C)−1 for Nichrome

To solve this, we would use the formula

P = V² / R

So when we rearrange and make R subject of formula, we have

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 ohms

Recall the formula for resistivity of a wire

R = ρ.L/A

Again, in rearranging and making L subject of formula, we have

L = R.A / ρ

To make it uniform, we convert our radius from mm to m.

Diameter, D = 0.5 mm

Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m

We then use this radius to find our area

A = πr²

A = π * 0.00025²

A = 1.96*10^-7 m²

And finally, we solve for L

L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =

L = 3.162 m

(b)

Temperature coefficient of resistance.

R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]

R₁₂₀₀ = R₂₀[1 + α(1180) ]

R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]

R₁₂₀₀ = 24.2[1 + 0.472]

R₁₂₀₀ = 24.2 * 1.472

R₁₂₀₀ = 35.62 ohms

Putting this value of R in the first formula from part a, we have

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Which of these objects are constantly in motion? Select all that apply.A.
Earth

B.
Planes

C.
Trains

D.
Blood

E.
Sun

F.
Cars

Answers

The object Earth,Sun, and Blood are constantly in motion. The correct option is A, D, and E.

What is motion?

if a body changes its position with respect to its surroundings in a given interval of time, Then the body is said to be in motion

Motion is generally classified as follows.:

i) Rectilinear motion.

ii) Circular motion.

iii) Rotational motion.

iv) Periodic motion.

The Earth is continuously in motion because it continuously revolves around The Sun in an elliptical orbit, due to which a year is 365 days. and also The earth rotates about its own axis once a day.

The Sun also revolves around the galactic center of our Milkyway galaxy. and it also rotates about its own axis continuously. so that the sun is also continuously in motion.

The Human blood is continuously in motion Because our blood is continuously circulating whole over the body with the help of our heart. The heart continuously pumps our blood and circulates it inside the human body.

Hence the Earth, Sun, and blood are continuously in motion.

To learn more about Motion click:

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#SPJ3

earth, blood, and the sun

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for F⃗ ? Let U=0 when x=0. Express your answer in terms of α and x.

Answers

Answer

$\Delta U= \alpha (x^3)/(3) \n$

Explanation:

given

$F = -\alpha x^2 i$

where $\alpha = 12 N/m^2$

now we know

$\int\limits^W_0 {} \, dW = \int\limits^a_b {F.} \, dxi$ ..................(i)

where dx is infinitesimal distance

$W = \int\limits^a_b {-\alpha x^2} \, dx \n$

for x = a and b = 0

after integration we get

$W = -\alpha (x^3)/(3)$

we know work done by conservative force will be equals to negative of potential energy

$W = -\Delta U$

so we get

$-\Delta U= -\alpha (x^3)/(3) \n\n\Delta U= \alpha (x^3)/(3) \n$

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has a magnitude of 0.250 T and direction from east to west.A)If air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.

Use 1.602×10−19 C for the magnitude of the charge on an electron.

B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.

a-from north to south

b-from south to north

Answers

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

$v_(f)^(2)=v_(0)^(2)+2gy\nv_(f)=\sqrt{0+2(9.8(m)/(s^(2)))(145m)}=53.31(m)/(s)$

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

$F=(3.8*10^(8))(1.602*10^(-19)C)(53.31(m)/(s))(0.205T)=6.65*10^(-10)N$

b. south - north (by the rigth hand rule)

I hope this is usefull for you

regards

2. Annealed low-carbon steel has a flow curve with strength coefficient of 75000 psi and strain-hardening exponent of 0.25. A tensile test specimen with a gauge length of 2 in. is stretched to a length of 3.3 in. Determine the flow stress and the average flow stress that the metal experienced during this deformation.

Answers

Answer:

Flow stress= 9390Psi

Average flow stress= 4173.33Psi

Explanation:

Given:

Strength Coefficient = 75000psi

Strain hardening Exponent = 0.25

Gauge length = 2inches

Stretch length = 3.3 inches

Flow stress,Yf = 75000 × ln(3.3/2) × 0.25

Yf = 75000× ln(1.65) × 0.25

Yf = 75000× 0.5008 × 0.25

Flow stress = 9390Psi

Average flow stress = 75000× 0.5008 × (0.25/2.25)

Average flow stress= 4173.33psi

When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

Answers

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

mass of the box, m = 5 kg
extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

$k = (mg)/(x) \n\nk = (5 * 9.8)/(0.05) \n\nk = 980 \ N/m$

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

$T = kx\n\nx = (T)/(k) \n\nx = (59)/(980) \n\nx = 0.06 \ m\n\nx = 60 \ mm$

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

For free fall, a = g

T = m(g - a) = 0

Learn more here:brainly.com/question/4404276

Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

we know that;

mg = kX

5 × 9.81 = k(0.05)

k = 981 N/m

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

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