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olga nikolaevna [1]
3 years ago
5

A 2.3 kg block executes SHM while attached to a horizontal spring of spring constant 150 N/m. The maximum speed of the block as

it slides on a horizontal frictionless surface is 2.3 m/s. What are (a) the amplitude of the block's motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete 4.1 cycles of its motion?
Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer

given,

mass of block = m = 2.3 Kg

spring constant = k = 150 N/m

speed = 2.3 m/s

a) we know,

   \omega = \sqrt{\dfrac{k}{m}}

   \omega = \sqrt{\dfrac{150}{2.3}}

             ω = 8.08 rad/s

      v = A ω

      2.3 = A x 8.08

         A = 0.285 m

b) maximum acceleration

        a = A ω²

        a = 0.285 x 8.08²

        a = 18.58 m/s²

c) acceleration is minimum at mean position the minimum value of acceleration is zero.

d) time period

   T = \dfrac{2\pi}{\omega}

   T = \dfrac{2\pi}{8.08}

            T = 0.778 s

        t = 4.1 x 0.7778

        t = 3.188 s

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VikaD [51]

Answer:

The answer is A.

Explanation:

Hypertrophy is an increase and growth of muscle cells. Hypertrophy refers to an increase in muscular size achieved through exercise. When you work out, if you want to tone or improve muscle definition, lifting weights is the most common way to increase hypertrophy.

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5 0
2 years ago
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A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
3 years ago
70 POINTS AND BRAINLIEST PLEASE HELP!!!The Moon orbits Earth. This orbit causes the Moon to look different over the course of ab
Rudik [331]

Answer:

It always acurs after a 1st quarter

do you have photo?

Explanation:

8 0
3 years ago
A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if
tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\

Fm is the moving force

\mu is  the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is  4.16m/s²

4 0
2 years ago
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th
Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

  • A uniform motion along the horizontal (x) direction
  • A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

s=u_y t + \frac{1}{2}at^2

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

where

u = 23.5 m/s is the initial velocity

\theta=33.5^{\circ} is the angle of projection

Substituting,

u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

Substituting everything into the equation we get:

-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s

Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0
3 years ago
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