Balls and rods set up to represent a solar system

Oliga [24]

Answer:

student attach a save block to a horizontal spring so that the block spring system will oscillator with the block spring system released from rest horizontal position that is not the systems equilibrium position well this question regards about the energy used the answer may be 0.73 Joel ok you just try it ok verified

Explanation:

apply applied the potential energy value mean the formula MGH write it means what mass into gravitation in to height

6 0

3 years ago

How much heat would be needed to completely evaporate 31.5 g of boiling water at a temperature of 100 "C? Express your answer in

BaLLatris [955]

Answer:

Heat needed = 71.19 J

Explanation:

Here heat required can be calculated by the formula

H = mL

M is the mass of water and L is the latent heat of vaporization.

Mass of water, m = 31.5 g = 0.0315 kg

Latent heat of vaporization of water = 2260 kJ/kg

Substituting

H = mL = 0.0315 x 2260 = 71.19 kJ

Heat needed = 71.19 J

7 0

3 years ago

Find the intensity in decibels [i(db)] for each value of i. normal conversation: i = 106i0 i(db) = power saw a 3 feet: i = 1011i

White raven [17]

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

8 0

3 years ago

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A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

4 years ago

Use the scale drawing to find the distance represented by AB

Andre45 [30]

The answer to your question is AB is 50 kilometers.

50 kilometers

5 0

3 years ago

Read 2 more answers