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2 years ago

9

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

lly compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.

1 answer:

BabaBlast [244]2 years ago

5 0

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

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The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

$F_1 = \frac{kq_1q_2}{r_1^2}$ --- Coulomb's law

Substitute 2 for r1

$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

$F_2 = \frac{F_1}{4}$

The new force is 1/4 of the previous force.

3 0

2 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

2 years ago

Suppose that the inverse market demand for silicone replacement tips for Sony earbud headphones is p = pN - 0.1Q, where p is t

lbvjy [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The effect of a change in the price of a new pair of headphones on the equilibrium price of replacement tips ( dp/dpN) is

$\frac{\delta p}{\delta p_N} =1$

b

The value of Q and p at equilibruim is

$Q_e = 250$ and $p_d =$ 5

The consumer surplus is $C= 3125$

The producer surplus is $P = 375$

Explanation:

From the question we are told that

The inverse market demand is $p_d = p_N -0.1Q$

The inverse supply function is $p_s = 2+ 0.012Q$

a

The effect of change in the price is mathematically given as

$\frac{\delta p}{\delta p_N}$

Now differntiating the inverse market demand function with respect to $p_N$

We get that

$\frac{\delta p}{\delta p_N} =1$

b

We are told that $p_N =$ $30

Therefore the inverse market demand becomes

$p_d = 30 -0.1Q$

At equilibrium

$p_d = p_s$

So we have

$30 -0.1Q_e = 2+ 0.012Q_e$

Where $Q_e$ is the quantity at equilibrium

$28 = 0.112Q_e$

$Q_e = \frac{28}{0.112}$

$Q_e = 250$

Substituting the value of Q into the equation for the inverse market demand function

$p_d = 30 - 0.1 (250 )$

$p_d =$ 5

Looking at the equation for $p_d \ and \ p_s$ we see that

For Q = 0

$p_d = 30$

$p_s =2$

And for Q = 250

$p_d = 5$

$p_s = 5$

Hence the consumer surplus is mathematically evaluated as

$C = \frac{1}{2} * Q_e * (30 -5)$

Substituting value

$C = \frac{1}{2} * 250 (30-5)$

$C= 3125$

And

The producer surplus is mathematically evaluated as

$P = \frac{1}{2} *250 * (5-2)$

$P = 375$

4 0

2 years ago

Mention two ways to increase the strength of an electro magnet.

VikaD [51]

1.) Law of Induction.

2.) Increase Winding Count

6 0

3 years ago

Read 2 more answers