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3 years ago

what is the maximum speed of a 34-g object bouncing on a spring (k=78.1 N/m) with an amplitude of 3.5-cm?

Physics

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

The maximum speed of the mass is 1.67 m/s.

Explanation:

We have,

Mass of object is 34 g or 0.034 kg

Spring constant of the spring is 78.1 N/m

Amplitude attained by the object is 3.5 cm or 0.035 m

It is required to find the maximum speed of the object in this spring mass system. The maximum speed is given by :

$v=A\omega$

$\omega=\sqrt{\dfrac{k}{m}}$

$v=A\sqrt{\dfrac{k}{m}}$

Plugging all the values in above formula,

$v=0.035\times \sqrt{\dfrac{78.1}{0.034}}\\\\v=1.67\ m/s$

So, the maximum speed of the mass is 1.67 m/s.

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Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is

Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0

3 years ago

Read 2 more answers

You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5 + 0.6 × 20c = 343.5

so wavelength of sound λ = c / f

wavelength = 343.5 / 686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............

for d = 0.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)$ = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)$ = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)$ = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475 = 2.975 m

distance 2 = 2.5 + 3.125 = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0

4 years ago

Electron configuration is responsible for the make up of the chemical properties of elements

allochka39001 [22]

This is True

I hope i helped

3 0

3 years ago

Read 2 more answers

A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate

qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg = 0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0

3 years ago

Strontium has density of 2.64 g/cm3 and crystallizes with the face-centered cubic unit cell. Calculate the radius of a strontium

pantera1 [17]

Answer:

The radius of strontium atom is

2.14 E-8cm

Explanation:

Go through the attached file for a comprehensive detailed explanation.

6 0

3 years ago