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2 years ago

Solve A and B using energy considerations.

1 answer:

7 0

Use the kinematic equation vf^2 = vi^2 + 2ad where;
vf = ?
vi = 0 m/s
a = 9.8 m/s^2
d1 = 10 m
d2 = 25 m

final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s

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Betty weighs 420 n and she is sitting on a playground swing seat that hangs 0.40 m above the ground. tom pulls the swing back an

Svet_ta [14]

Weight = mass * gravity
420 = mass * 9.8
mass of Betty = 42.857 kg

Difference in height = 1 - 0.45 = 0.55 meters

Total energy = Kinetic energy + potential energy

At the highest point, the kinetic energy is zero while the potential energy is maximum, therefore, we can get the total energy as follows:
Total energy = 0 + mgh
Total energy = 42.857*9.8*0.55 = 231 Joules

At the lowest point, the potential energy is zero while the kinetic energy is maximum. Therefore:
Total energy = 0.5 * m * (v)^2 + 0
231 = 0.5 * (42.857) * (velocity)^2
(velocity)^2 = 10.78
velocity = 3.28 meters/sec

8 0

2 years ago

A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in

stepladder [879]

Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s

F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)

F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
F = -47,222 N The negative sign means that the force vector is 
applied AGAINST the momentum vector of the rhinoceros.

7 0

2 years ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0

2 years ago

Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.

denis23 [38]

Answer:

a) We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength

b) There is resonance for the lengths 25 and 75 cm

c) Resonance occurs for tubes with length 31 and 93 cm

Explanation:

To find the length of the tube that has resonance we must find the natural frequencies of the tubes, for this at the point that the tube is closed we have a node and the open point we have a belly; in this case the fundamental wave is

λ = 4L

The next resonance called first harmonic λ₃ = 4L / 3

The next fifth harmonic resonance λ₅ = 4L / 5,

WE see that the general form is λ ₙ= 4L / n n = 1, 3, 5 ...

Let's use these expressions for our problem

Let's start with the shortest wavelength.

a) Lam = 60 cm

Let's look for the tube length that this harmonica gives

L = λ n / 4

To find the shortest tube length n = 1

L = 60 1/4

L = 15 cm

For n = 3

L = 60 3/4

L = 45 cm

For n = 5

L = 60 5/4

L = 75 cm

For n = 7

L = 60 7/4

L = 105cm

We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength, in different harmonics 1, 3 and 5

.b) λ = 100 cm

For n = 1

L = 100 1/4

L = 25 cm

For n = 3

L = 100 3/4

L = 75 cm

For n = 5

L = 100 5/4

L = 125 cm

There is resonance for the lengths 25 and 75 cm in the fundamental and third ammonium frequency

c) λ= 124 cm

L = 124 1/4

L = 31 cm

For the second resonance

L = 124 3/4

L = 93 cm

Resonance occurs for tubes with length 31 and 93 cm in the fundamental harmonics and third harmonics

8 0

1 year ago

A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at

mote1985 [20]

Answer:

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

8 0

1 year ago