How much time should be allowed for a 605-mile car trip if the car will the traveling at an average speed of 55 Miles per hour?

A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the

iren2701 [21]

Answer:

a) T = 1,467 s , b) A = 0.495 m , c) v = 4.97 10⁻² m / s

Explanation:

The simple harmonic movement is described by the expression

x = A cos (wt + Ф)

Where the angular velocity is

w = √ k / m

a) Ask the period

Angular velocity, frequency and period are related

w = 2π f = 2π / T

T = 2π / w

T = 2pi √ m / k

T = 2π √ (1.2 / 22)

T = 1,467 s

f = 1 / T

f = 0.68 Hz

b) ask the amplitude

The mechanical energy of a harmonic oscillator

E = ½ k A²

A = √2 E / k

A = √ (2 2.7 / 22)

A = 0.495 m

c) the mass changes to 8.0 kg

As released from rest Ф = 0, the equation remains

x = A cos wt

w = √ (22/8)

w = 1,658

x = 3.0 cos (1,658 t)

Speed is

v = dx / dt

v = -A w sin wt

The speed is maximum when without wt = ±1

v = Aw

v = 0.03 1,658

v = 4.97 10⁻² m / s

6 0

3 years ago

A well is being dug. A 4.5-kg bucket is filled with 28.0 kg of dirt and pulled vertically upward at a constant speed through a d

andreyandreev [35.5K]

The work done on the filled bucket in raising out of the hole is 2, 925 Joules

<h3>How to determine the work done</h3>

Using the formula:

Work done = force * distance

Note that force = mass * acceleration

F = mg + ma

F = 4. 5 * 10 + 28 * 10

F = 45 + 280

F = 325 Newton

Distance = 9m

Substitute into formula

Work done = 325 * 9

Work done = 2, 925 Joules

Therefore, the work done is 2, 925 Joules

Learn more about work done here:

brainly.com/question/25573309

#SPJ1

4 0

2 years ago

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

If V is the original volume, V' is the new volume, T is the original Kelvin temperature, and T' is the new Kelvin temperature, h

madam [21]

Answer:

$\frac{V}{T} = \frac{V'}{T'}$

Explanation:

Given the following data;

Original volume = V

New volume = V'

Original temperature = T

New temperature = T'

To write an expression for Charles's law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles law is given by the formula;

$\frac {V}{T} = K$

$\frac{V}{T} = \frac{V'}{T'}$

4 0

3 years ago