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alexgriva [62]
2 years ago
6

Please help me with this question I’m honestly done with this class.

Physics
1 answer:
maw [93]2 years ago
3 0

Answer: when it is stretched as far as possible

Explanation:

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A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sin t 6 N (newtons) and moves in a medi
nordsb [41]

Answer:

u" + 40u' + 49u = 2 sin(t/6)        

upp + 40up + 49u = 2 sin(t/6)

Explanation:

Step 1: Data given

mass = 5 kg

L = 20 cm = 0.2 m

F = 10 sin(t/6)N

Fd(t) = - 6 N

u(0) = 0.03 m/s

u(0) = 0

u'(0) = 3 cm/s

Step 2:

ω =kL

k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²

Since Fd(t) = -γu'(t)  we know:

γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m

The initial value problem which describes the motion of the mass is given by

5u" + 200u' + 245u = 10 sin(t/6)   u(0) = 0  ;  u'(0) = 0.03

This is equivalent to:

u" + 40u' + 49u = 2 sin(t/6)        u(0) = 0  ;  u'(0) = 0.03

upp + 40up + 49u = 2 sin(t/6)

With u in m and t in s

5 0
3 years ago
A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

3 0
3 years ago
In Which senecio is an animal doing work
Arisa [49]

Answer:

A horse pulls a wagon along a road

4 0
3 years ago
g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass m
marta [7]

Answer: hello question b is incomplete attached below is the missing question

a) attached below

b) V = 0.336 ft/s

Explanation:

Elongation ( Xo)  = 16/ 7 feet

mass attached to 4-foot spring = 16 pounds

medium has 9/2 times instanteous velocity

<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>

The motion is an underdamped motion because the value of β < Wo

Wo = 3.741 s^-1

attached below is a detailed solution of the question

3 0
2 years ago
How are evaporation and transpiration similar?
Iteru [2.4K]
The answer is A. They are both processes in which water is changed into water vapor.
7 0
3 years ago
Read 2 more answers
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