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

Answer:

vestigation 4

Explanation:

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.

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

Answer:

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

These labels mean that each lightbulb has Its respective power delivered to It when It Is connected to a constant

Explanation:

Answer:

c

Explanation:

Answer:

The energy stored is: 62.5 Joules

Explanation:

Given

[tex]k = 500N/m[/tex] --- spring constant

[tex]x = 0.5m[/tex] --- stretch

Required

The amount of energy

This is calculated as:

[tex]U = \frac{1}{2} kx^2[/tex]

[tex]U = \frac{1}{2} * 500N/m * (0.5m)^2[/tex]

[tex]U = 250N/m * (0.5m)^2[/tex]

[tex]U = 62.5\ J[/tex]

Answer:

It transfers energy parallel to the direction of wave motion. A transverse wave..

Explanation:

Answer:

The answer is a Wave Motion

Explanation:

A longitudinal wave is a type of wave that transfers energy is called," Wave Motion".

Hope you have a great day :)

Answer:

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

Answer:

X-rays have so much energy they can go right through your muscles and fat.

Explanation:

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.

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.

Answer:

a

Explanation:

mark me as brainlist ez

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 )

    = ( 1/2 * 17459.18 * 0.50^2) + ( 17459.18 * 9.81 * 120 )

    = 2.053 * 10^7 J

Answer: lollllllllllllllllllllll

Explanation:

Answer:

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

Explanation:

Answer:

true hope this help :}

Explanation:

Answer:

Magma from the core

Explanation:

...?

Answer:

[tex]343 \div 1000 = 0.343[/tex]

wavelength is equivalent to the velocity divided by the frequency

Answer:

work=f×d

GPE =m×g×h

[tex]kinetic energy = \frac{1}{2} mv {}^{2} \\ power = \frac{w}{t} [/tex]

Answer: 88

Explanation:

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

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:

what they said

Explanation:

Answer:

Explanation:

Kamal, Kamala, Krishna invert  in the ratio 2 : 3 : 5.

As 2+3+5 =10, their respective share is 2/10, 3/10 and 5/10.

Sum of investment = 30000Rs

Kamal's investment = 30000 * 2/10 = 6,000 Rs

Kamala's investment = 30000 * 3/10 = 9,000 Rs

Krishna's investment = 30000 * 5/10 = 15,000 Rs

Answer:

yes it is a reactant

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

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

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

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