The work-energy theorem states that the work done on an object is equal to a change in which quantity?

Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks

babymother [125]

Answer:

a) The answer is 11,7m

b) The time it takes to fall will be shorter

Explanation:

We will use the next semi-parabolic movement equations

$H=Hi+Viy*t+1/2*g*t$

$X=Xi+Vx*t$

Where g(gravity acceleration)=9,81m/s^2

Also Xi, Hi and Viy are zero, as the stones Billy-Jones is kicking stay still before he moves them, so we take that point as the reference point

The first we must do is to find how much time the stones take to fall, this way:

$t=(5.40m)/(3.50m/s)$

Then t=1,54s

After that we need to replace t to find H, this way

$H=(1/2)*(9,81m/s^2)*(1,54s)^2$

Then H=11,7m

b) The stones will fall faster as the stones will be kicked harder, it will cause the stones move faster, it means, more horizontal velocity. In order to see it better we could assume the actual velocity is two times more than it is, so it will give us half of the time, this way:

$t=(5,40m)/(2*3,50m/s)$

Then, t=0,77

4 0

3 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0

3 years ago

What would be the answer for this and how?

Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

$k = \frac{F}{\Delta x}\\$

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

$k = \frac{10\ N}{0.04\ m}\\k = 250\ N/m\\$

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

$k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\$

For a load of 30 N, applying Hooke's Law:

$\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\$

Hence, the correct option is:

7 0

2 years ago

Which units are used to measure both velocity and speed? Select three options.

xz_007 [3.2K]

M/s, km/h, and mph are all used to measure these quantities

6 0

3 years ago

Read 2 more answers

A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the

Serjik [45]

The elastic potential energy of a spring is given by
$U= \frac{1}{2}kx^2$
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
$W=-\Delta U = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$

In our problem, initially the spring is uncompressed, so $x_i=0$ . Therefore, the work done by the spring when it is compressed until $x_f$ is
$W=- \frac{1}{2}kx_f^2$
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.

8 0

3 years ago