Physics High School

A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point foot above the equilibrium position with a downward velocity of ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant

Answers

Answer 1

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c) 3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity hence β = 1

a)Calculate the time at which the mass passes through the equilibrium position.

time mass passes through equilibrium = 1/12 seconds = 0.083

b) Calculate the time at which the mass attains its extreme displacement

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

c) What is the position of the mass at this instant

position = 3e^-4/3

attached below is the detailed solution to the given problem

A Mass Weighing Pounds Is Attached To A Spring Whose Constant Is Lb/ft. The Medium Offers A Damping Force

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What is the power of a refrigerator with voltage 110 V and
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Answer: 88

Explanation:

First term / second mid Exam [1433/ 1434] Q12. A conductor of radius I, length and resistivity p has resistance R. What is the new resistance if it is stretched to 4 times its original length?

Answers

Answer:

R' = 4R

Hence, the resistance increases to four times its initial value as well.

Explanation:

The resistance of a conductor in terms of its physical dimensions can be given by the following formula:

[tex]R = \frac{\rho L}{A}\\[/tex] _______ equation (1)

where,

R = Resistance of the conductor

ρ = resistivity of the material of the conductor

L = Length of the conductor

A = Cross-sectional Area of the conductor

Now, if the length of the conductor is increased to four times, keeping other values constant, then:

[tex]R' = \frac{\rho L'}{A}\\\\R' = \frac{\rho (4L)}{A}\\\\R' = 4\frac{\rho L}{A}[/tex]

using equation (1):

R' = 4R

Hence, the resistance increases to four times its initial value as well.

Please please please please please please please please please please please please please please please please please please please

Answers

c i think & mark as brain list?

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his acceleration

Answers

Answer:

1 m/s²

Explanation:

From the question,

Using

a = (v-u)/t.................... Equation 1

Where a = accelartion of the bicycle, v = Final velocity, u = initial velocity, t = time.

Given: v = 15 m/s, u = 5 m/s, t = 10 s

Substitute these values into equation 1

a = (15-5)/10

a = 10/10

a = 1 m/s²

Hence the acceleration of the bicycle is 1 m/s²

1. A +4 nC charge is placed at a distance r from another charge. If the charge experiences a force of 6000 N, what is the electric field intensity E ?

Answers

Answer:

1500 N/C

Explanation:

From the question,

Applying,

E = F/q.................. Equation 1

Where E = Electric Field intensity, F = Force experienced by the charge, q = Charge

Given: F = 6000 N, q = +4nC

Substitute these value into equation 1

E = 600/4

E = 6000/4

E = 1500 N/C

Hence, the electric field intensity is 1500 N/C.

Which material transmits the most visible light?
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B.Mirror
C.Clear glass
D.Black cloth

Answers

Answer:

Explanation:

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his acceleration

Answers

[tex]\Large {{ \sf {Question :}}} [/tex]

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?

[tex]\Large {{ \sf {Given :}}} [/tex]

Initial Velocity (u) - 5 m/sFinal Velocity (v) - 15 m/sTime (t) - 10 sec

[tex]\Large {{ \sf {Formulae :}}} [/tex]

If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is, [tex]a \: = \frac{v - u}{t} [/tex][tex]\Large {{ \sf {Step-by-step explanation :}}} [/tex]

[tex]a \: = \frac{v - u}{t} \\ or \: \: a = \frac{(15 - 5)}{10} m \: s^{ - 2} \\ or \: \: a \: = \frac{10}{10}m \: s^{ - 2} \\ or \: \: a = 1m \: s^{ - 2} [/tex]

[tex]\Large {{ \sf {Answer :}}} [/tex]

His acceleration is [tex]1m \: s^{ - 2} [/tex]

ممكن حل بسرعة
please help

Answers

Answer:

I could help but i dont speak or understand a word of Arabic im so sorry.

Explanation:

What are the purposes of these items aboard the Voyager probes? • Low-energy particle detector: • Magnetometer boom: • High-gain antenna: • Photopolarimeter: • Cosmic ray:

Answers

Explanation:

1. Low-energy particle detector: This particle detector measures the charged particles of the solar winds.

2. Magnetometer boom: This device measures magnetic fields produced by astronomical bodies.

3. High-gain antenna: A HGA has a narrow radio beam that is used to enhance the strength of signal. They simply amplify the weak signals.

4. Photopolarimeter: This is an instrument that is used to measure the strength and intensity and polarization of reflected light.

Answer:

Answer:

work=f×d

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In the 1500's Galileo experimented and discovered many things. One of his famous experiments allowed him to discover that blank when you drop them.

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Answer:

In the 1500's Galileo experimented and discovered many things. One of his famous experiments allowed him to discover that all Objects fall the same rate Bolo when you drop them.

Explanation:

for mo

PLEASE HELP! A 0.423 kg object carries a +11.5 uc
charge. It is 0.925 m from a -7.55 C
charge. What is the magnitude of the
object's acceleration?
(u stands for micro. Remember, magnitudes
are positive.)
[?] m/s2

Answers

Answer:

2.15 m/s²

Explanation:

We'll begin by calculating the force of attraction between two charges. This can be obtained as follow:

Charge of 1st object (q₁) = +11.5 μC = +11.5×10¯⁶ C

Charge of 2nd object (q₂) = –7.55 μC = –7.55×10¯⁶ C

Electrical constant (K) = 9×10⁹ Nm²/C²

Distance apart (r) = 0.925 m

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 11.5×10¯⁶ × 7.55×10¯⁶/ 0.925²

F = 0.781425 / 0.855625

F = 0.91 N

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass of object (m) = 0.423 Kg

Force (F) = 0.91 N

Acceleration (a) =?

F = ma

0.91 = 0.423 × a

Divide both side by 0.423

a = 0.91 / 0.423

a = 2.15 m/s²

Thus, the magnitude of the object's acceleration is 2.15 m/s²

Answer:

2.15

Explanation:

Acellus

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Answers

Answer:

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Explanation:

The answer is true for this

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Answers

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Answers

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Group of answer choices

A 12 N of force pulling left while a 22 N of force pulls right.

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A 2 N force and a 10 N force pulling left while a 12 N force pulls right.

A 22 N force pulling left while a 6 N force and a 12 N force pulls right.

Answers

Answer:

A 2 N force pulling and a 10 N force pulling left while a 12N force pulls right.

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There are actually more ways to fill in those two blanks... Different parts of the body absorb x-rays in varying degrees, but soft tissues (like skin, muscles, fat, and organs) allow most of the X-rays to pass through.

A lightbulb has 25 W stamped on it. What does this mean?

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Answer:

A lightbulb has 25 W stamped on it. What does this mean?:

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A cog system on the beginning segment of a roller coaster needs to get 29 occupied cars up a 120-m vertical rise over a time interval of 60 s. Each car experiences a gravitational force of 5900 N. The cars start at rest and end up moving at 0.50 m/s.

Answers

Answer:

Hello your question has some missing part attached below is the missing part

How much work is done by the cog system on the cars?

answer : 2.053 * 10^7 J

Explanation:

Total weight of car ( wT ) = 5900 * 29 = 171100 N

mass of car ( mT ) = wT / g = 171100 / 9.81 = 17459.18 kg

Hence work done by the cog system on the cars

W = ( KE[tex]_{2}[/tex] + PE[tex]_{2}[/tex] ) - ( KE[tex]_{1}[/tex] + PE[tex]_{1}[/tex] )

= ( 1/2 mv^2 + mgh ) - ( 0 )

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= 2.053 * 10^7 J

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