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3 years ago

11

. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42

]. Find, a) the time for one complete vibration; b) the force constant of the spring; c) the maximum speed of the mass; d) the maximum force on the mass; e) at t= 1.00s, find the mass is a. the position, b. the speed, c. the acceleration d. the force on the mass

1 answer:

rusak2 [61]3 years ago

5 0

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

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3 years ago

Amanda [17]

Answer:

When a particle or a system of particles move in a system where no external force acts, then the total linear momentum of the particle system remains constant.

Explanation:

Given data:

Total mass of the skateboarder, $m_{1} = 70 \ kg$

Mass of the friend, $m_{2} = 50 \ kg$

Initial velocity of the skateboarder, $u_{1} = 3 \ m/s$

Initial velocity of the the friend, $u_{2} = 0$

Let the new velocity of the skateboarder when his friend jumps be $v$ .

From the conservation law of linear momentum,

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

$70 \times 3 + 50 \times 0 = (70 + 50)v$

$\Rightarrow \ v &= 1.75 \ m/s.$

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4 years ago

Why do storm go to Galveston?

notsponge [240]

Hurricanes and tropical storms can also spawn tornadoes and microbursts, create storm surges ... If you are unable to evacuate, go to your wind-safe room.

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4 years ago

Predict whether sound will would travel faster in air in summer or in winter?

Arisa [49]

In summer because the air in summer is less dense.

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3 years ago

What acceleration will you give to a 25kg box if you push it with a force of 85N

Grace [21]

<h3>Answer:</h3>

3.4 m/s²

<h3>Explanation:</h3>

We are given;

Mass of the box as 25 kg
Force is 85 N

We are required to determine the acceleration;

According to second newton's law of motion force is given by the product of mass and acceleration.
That is;

Force = ma

Rearranging the formula;

a = F ÷ m

Therefore;

acceleration = 85 N ÷ 25 kg

= 3.4 m/s²

Thus, the acceleration of the box will be 3.4 m/s²

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