Does a 2000 mercury cougar have a timing belt

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

1. A sample of gas has a constant temperature and number of particles. As the volume of the gas sample is increased, the pressur

AURORKA [14]

Answer:

1.C

2.C

3.C

Explanation:

hope its help hehe:(

4 0

3 years ago

Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an

nadezda [96]

Answer:

E+mc+UD-3

Explanation:

7 0

4 years ago

Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro

ehidna [41]

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

$T=2\pi\sqrt{\frac{l}{g} }$

angular frequency ω = 2π / T

= $\omega=\sqrt{\frac{g}{l} }$

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 ( ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

4 0

4 years ago

Earth's magnetic field is approximately 1/2 gauss, that is 50 micro-tesla because the SI field unit of a tesla is 10,000 gauss.

jek_recluse [69]

Answer:

F = 0

Explanation:

The magnetic force is described by two expressions

for a moving charge

F = q v x B

for a wire with a current

F = I L xB

bold indicates vectors

let's write this equation in module form

F = I L B sin θ

where the angle is between the direction of the current and the direction of the magnetic field

In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current

The magnetic field of the Earth goes from the south to the north and in this part it is horizontal

Therefore the current and the magnetic field are parallel, the angle between them is zero

sin 0 = 0

consequently the magnetic force is zero

F = 0

5 0

3 years ago