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leva [86]
3 years ago
11

a 2.00 kg friction-less block is attached to an ideal spring with force constant 315 N/m.Initially, the spring is neither stretc

hed nor compressed , but the block is moving in the negative direction at 12.0 m/s. find:-a) the amplitude.b) the maximum acceleration of the block.c) the maximum force the spring exerts on the block.
Physics
1 answer:
Setler [38]3 years ago
5 0

Answer

given,

mass of block, m = 2 Kg

spring constant, k = 315 N/m

speed of the block, v = 12 m/s

a) Amplitude of the motion

   A = v \sqrt{\dfrac{k}{m}}

   A =\dfrac{12}{\sqrt{\dfrac{315}{2}}}

         A = 0.956 m

b) maximum acceleration of the block

    a_{max}= A \dfrac{k}{m}

    a_{max}= 0.956\times \dfrac{315}{2}

    a_{max}= 150.57\ m/s^2

c) maximum Force

  F_{max} = m a_{max}

  F_{max} = 2\times 150.57

  F_{max} = 301.14\ N

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3 years ago
Gauss's law is usualy written as :
snow_tiger [21]

Answer:

(a) the net charge inside the closed surface.

Explanation:

In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.

4 0
3 years ago
Que fuerza se obtendrá en el émbolo mayor de una prensa hidraúlica cuya área es de 167 cm2, cuando el émbolo menor de área igual
stiv31 [10]

Responder:

<h2>5.368N, </h2>

Explicación:

Según el principio pascal, establece que la presión aplicada en un punto sobre un líquido en un recipiente cerrado es igual a igual a la presión en cualquier otro punto del líquido.

Matemáticamente Presión ejercida por el pistón más pequeño = Presión ejercida por el pistón más grande.

La presión es la relación entre la fuerza y ​​su área de sección transversal.

P = Fuerza / Área de sección transversal

Sea P1 la presión sobre el pistón más pequeño y P2 la presión ejercida por el pistón más grande.

Como P1 = P2 entonces;

F1 / A1 = F2 / A2

Dado F1 = 450N, A1 = 14 cm², A2 = 167 cm² y F2 =?

Sustituyendo el valor conocido en la fórmula para obtener el requerido, tenemos;

\frac{450}{14} = \frac{F2}{167}\\  \\Cross\ multiplying\\\\14*F_2 = 450*167\\\\14F_2 = 75,150\\\\F_2 = \frac{75,150}{14} \\\\F_2 = 5,367.9N

F_2 \approx 5, 368N

Por lo tanto, la fuerza que se obtendrá en el pistón más grande de una prensa hidráulica cuya área es de 167 cm² es aproximadamente 5,368N,

4 0
3 years ago
A tennis player tosses a tennis ball straight up and then catches it after 2.07 s at the same height as the point of release.
prisoha [69]

(a) The ball will accelerate at a speed of 9.81 m/s² while it is in flight.

(b) The ball will be moving at a speed of 0 m/sec when it reaches its highest point.

(c) The ball's initial upward velocity is 20.30 m/s.

(d) It will reach a maximum height of 21.0 m.

<h3 /><h3>What is speed?</h3>

Speed is defined as the change in distance with regard to time. A scalar quantity, speed. It is a temporal element, m/sec is its unit.

The information provided in the issue is;

u is the fall's beginning speed in m/sec;

h is the fall's distance

g is the fall's acceleration or 9.81 m/sec²

a. The ball's speed while in flight is equal to the acceleration caused by gravity, which is 9.81 m/s².

b. When the ball reaches its highest point, its velocity will be zero.

c. The ball's starting velocity is;

v=u-gt

0=u-gt

u=gt

u=9.81 m/s² × 2.07 sec

u=20.30 m/s

d. The formula is used to determine the highest height it can reach.

\rm u^2 = - 2gh \\\\ (20.30)^2 = -2 \times (-9.8) \times h \\\\ 412.09  = 19.6 \times h\\\\ h = 412.09/19.6\\\\ h = 21.00 m

As a result, the ball's greatest height, beginning velocity, and acceleration are all 9.81 meters per second, zero meters per second, 20.30 meters per second, and 21 meters accordingly.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

3 0
2 years ago
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0
3 years ago
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