Physics High School

A car bounces up and down on its springs at 1.0 Hz with only the driver in the car. Now the driver is joined by four friends. The new frequency of oscillation when the car bounces on its springs is

Answers

Answer 1

The new frequency of oscillation when the car bounces on its springs is 0.447 Hz

Frequency of oscillation of spring

The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where

k = spring constant and m = mass on spring

Now since k is constant, and f ∝ 1/√m.

So, we have f₂/f₁ = √(m₁/m₂) where

f₁ = initial frequency of spring = 1.0 Hz, m₁ = mass of driver, f₂ = final frequency of spring and m₂ = mass on spring when driver is joined by 4 friends = 5m₁

So, making f₂ subject of the formula, we have

f₂ = [√(m₁/m₂)]f₁

Substituting the values of the variables into the equation, we have

f₂ = [√(m₁/m₂)]f₁

f₂ = [√(m₁/5m₁)]1.0 Hz

f₂ = [√(1/5)]1.0 Hz

f₂ = 1.0 Hz/√5

f₂ = 1.0 Hz/2.236

f₂ = 0.447 Hz

So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz

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Related Questions

What force causes raindrops to fall to Earth?

A.lift
B.drag
C.gravity
D.buoyancy

Answers

Answer:

Gravity

Explanation:

Because gravity try to pull everything toward the earth center. So raindrops fall towards the ground

A small sphere with a charge of 4 nC is brought close to but does not touch uncharged conducting sphere A, which is grounded. The ground is disconnected, and the small sphere is removed. Sphere A is then brought close to but does not touch uncharged conducting sphere B. What is the net charge on sphere B as a result, and what is an explanation for the charge

Answers

Answer:

the answer is b that's my answer

Can an object have both potential and kinetic energy at the same time ?

Answers

Answer:

yes

Explanation:

An object can be elevated and moving at the same time, example: a thrown ball

Explanation:

Ofcourse! Objects can have both kinetic energy and potential energy at the same time.

An object can be elevated above the ground (have potential energy) and be moving at the same time (and also have kinetic energy).

8. A 55-kg skateboarder enters a ramp moving horizontally with a speed of 6.5 m/s, and leaves the ramp moving vertically with a speed of 4.1 m/s. (A) Find the height of the ramp, assuming no energy loss to frictional forces. (B) What is the skateboarder's maximum height above the bottom of the ramp?

Answers

(a) The energy lost to friction by the skateboarder is -699.6 J.

(b) The maximum height above the bottom of the ramp is 1.3 m.

Energy loss to friction

The energy lost to friction by the skateboarder is calculated from the change in kinetic energy of the skateboarder.

ΔK.E = K.Ef - K.Ei

ΔK.E = ¹/₂m(vi² - vf²)

ΔK.E = 0.5 x 55 (4.1² - 6.5²)

ΔK.E = -699.6 J

Maximum height above the bottom of the ramp

The maximum height above the bottom of the ramp is determined from conservation of energy.

mghi + ¹/₂mvi² = ¹/₂mvf² + mghf

55 x 9.8 x hi + (0.5 x 55 x 6.5²) = (0.5 x 55 x 4.1²) + (55 x 9.8 x 0)

539hi + 1161.88 = 462.28

539hi = -699.6

hi = -699.6/539

hi = -1.3 m

Thus, the maximum height above the bottom of the ramp is 1.3 m.

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A hummingbird beats its wings 80 times in 4 seconds. What is its period? (answer as a decimal, not a fraction.

What is the hummingbird's wingbeat frequency?

Answers

Answer:

20 per second

Explanation:

80/4

The hummingbird beats its wing 20 times per second!

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1. A Capacitor c =5.31 nF, is made of two Parallel square Plates seperated by distance d = 5mm. Find the Area. please ans this, i will give brainliest

Answers

We know that,

[tex]\huge ➸ C = ϵ₀ \frac{A}{d} [/tex]

C⤏Capacitanceϵ₀⤏permittivity of free space constantA⤏Area of capacitor platesd ⤏distance

given ↷

C⤏5.31 nF = 5.31 x 10^-9ϵ₀⤏8.85 x 10^-12 F/m A⤏Area of capacitor platesd ⤏5mm

to find ↷

The Area

solution ↷

[tex]\begin{gathered}➸ C = ϵ₀ \frac{A}{d} \\ \end{gathered} [/tex]

[tex]\begin{gathered}➸ A = \frac{Cd}{ϵ₀} \\ \end{gathered} \\ [/tex]

[tex]\begin{gathered}➸ A = \frac{5.31 \times {10}^{ - 9} \times 5 }{8.85 \times {10}^{ - 12} } \\ \end{gathered} \\ [/tex]

[tex]➸ A =3 \times {10}^{3} {mm}^{2} [/tex]

[tex]➸ A =3 {m}^{2} [/tex]

Hence, the area is 3m²

Answer:

Explanation:

The capacitance of two parallel square plates of area A and separated by distance d is given as:

C = ε(A/d) where ε is the permittivity of the dielectric material

For air, ε = 8.854x10^(-12) F/m

Substituting into the equation:

5.31x10^(-9) = 8.854x10^(-12) * A / 5x10^(-3)

A = 3.00 m^2

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