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3 years ago

A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Physics

1 answer:

Anastasy [175]3 years ago

7 0

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un

Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within 1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

$B_{i} =\frac{hr/2}{k}$

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

$\pi$ Dh(T-Tinf)= $I^{2} R_{e}$

make T the subject of the equation , we have

T=25+ $\frac{100^2*0.01}{\pi*0.001*500 }$

T=88.7 degC

rate of chnage in temperature

dT/dt= $\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)$

at t=o and integrating both sides $\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}$

we have

$\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}$

t=8.31s

steady state temperature =88.7deg C

t=time within 1 degC of it steady stae is 8.31s

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3 years ago

Where are warm air vents usually located in a forced air heating system

crimeas [40]

The system is located in the HairyFunyon aka near the floor

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3 years ago

Read 2 more answers

PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large contain

RSB [31]

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

Q_c = m ce ( $T_{f}$ -T₀)

Heat released by the largest container is

Q_a = M ce (T_{i}-T_{f})

how

Q_c = Q_a

m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process

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3 years ago

Jhon is 12 y o his train is 5 minutes late Jhon has an apple that wheigs half it s own mass calculate the mass of the sun

pav-90 [236]

Given the following information we have 20 watermelons from mark and 10 fishes from kim therefore we add the longitude of Walmart to the latitude of sams club and end up with a total of 1,000 dish soaps then we convert that into inches which leaves us at 20,000,000 inches of cats then multiply that number to 10 giraffes and we get
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3 years ago

A system initially has an internal energy U of 504 J. It undergoes a process during which it releases 111 J of heat energy to th

Likurg_2 [28]

Answer:

615 J

Explanation:

internal energy (U) = 504 J

heat lost (q) = 111 J = - 111 J (negative sign is because heat is lost)

work done = 222 J

what is the final energy in the system

total energy = final energy - initial energy

final energy = total energy + initial energy

where

initial energy = 504 J

total energy = 222 - 111 = 111 J

final energy = 504 + 111 = 615 J

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3 years ago