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3 years ago

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A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t

he impulse received by the ball from the floor during the bounce

1 answer:

irga5000 [103]3 years ago

6 0

Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

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Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area

yaroslaw [1]

Answer:

$5.63\cdot 10^{-6} C$

Explanation:

The capacitor of a parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

A is the area of each plate

d is the separation between the plates

$\epsilon_0$ is the vacuum permittivity

The energy stored in a capacitor instead is given by

$U=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

$U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}$

And re-arranging it

$Q=\sqrt{\frac{2U\epsilon_0 A}{d}}$

Now if we substitute

$d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J$

We find the charge stored on the capacitor:

$Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C$

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3 years ago

Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove

kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

$ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2$

$kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2$

Since the masses are equals:

$m=ma=mb$

For the known velocities, the kinetics energies result:

$ki=0.5*mVa_{i}^2$

$ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m$

$kf=0.5*mbVb_{f}^2$

$kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m$

The lost energy in the collision is the difference between the initial and final kinectic energies:

$kl=ki-kf$

$kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m$

Finally the relation between the lost and the initial kinetic energy:

$kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m$

$kl/ki = 24/49=0.49$

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3 years ago

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

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2 years ago

The strength of the gravitational pull between two object's is determined by?

iogann1982 [59]

Mass and distance are the two factors

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3 years ago

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A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage s

miskamm [114]

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

σ = n e2 τ / m*

The relationship with resistivity is

ρ = 1 /σ

Whereby the resistance

R = ρ L / A = 1 /σ L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

T = 2 T₀

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