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Elina [12.6K]
2 years ago
10

CHEGG In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75C as it moves at 0.2 m/s thro

ugh a straight thin-walled stainless steel tube of 12.7-mm diameter. A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the tube. If the tube is 10 m long, what is the required heat flux
Physics
1 answer:
vovangra [49]2 years ago
3 0

Answer:

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  • arris let lststdtsitsitlsitlsitl probably be perfect lstulitdults is a stlisitldudl us last leg e oil leak pal to Al pa pal rsrururuarul Al pa pal rsrururuarul lzulr para osutlsultsl to be pet lsu football game pa lslitsrulaulrslrusrykakryarulalutasulr ggggggggg is not me I will be there at the same time I don't have a car so I can get a ride to the airport fjztzhdilgdutldiydiltdultdutlstulzlurzykrUrlzultstlditldultsumtsultslutdliydilydlitdljdumtztulstzumtskyrUtlaraulszhfltulsfhlsjstultdisttilsdgkxidlutslutsutlzuzjzltisg
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Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing
Juliette [100K]

Answer: MR²

is the the moment of inertia  of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements  are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

3 0
3 years ago
A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring ​
PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

F=kx

where

F is the magnitude of the spring applied

k is the spring constant

x is the elongation of the spring, relative to its equilibrium position

For the spring in this problem, we have:

F = 0.12 N (force applied)

x = 3 cm = 0.03 m (elongation of the spring)

Therefore, we can solve the formula for k to find the spring constant:

k=\frac{F}{x}=\frac{0.12}{0.03}=4 N/m

Learn more about forces:

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4 0
3 years ago
A series circuit has a 12-volt power source and two resistors of 1 ohm and 2 ohms respectively. How many amps will flow when
Andrej [43]

Answer:

ayoooooooo

Explanation:

3 0
3 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.
jek_recluse [69]
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>
4 0
3 years ago
Read 2 more answers
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