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3 years ago

Stewart James, Calculus, Section 6.4, Page 449, Problem 5

Physics

1 answer:

faltersainse [42]3 years ago

4 0

The total work done is
<span>W=60 plus 120
=180J

For the graphs please find the attached image</span>

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A 500-kg golf cart is moving on level ground in a circular path of radius 10 m with a speed of 12 m/s. What is the centripetal f

Alborosie

Answer:

7200N

Explanation:

Centripetal force is directly proportional to the product of the mass and the square of the velocity and inversely proportional to the radius given.

8 0

2 years ago

Find your acceleration from 8.3 m/s to 12.5 m/s in 1.24 seconds.

balu736 [363]

The <u>average</u> acceleration for an object undergoing this change in velocity is

(12.5 m/s - 8.3 m/s) / (1.24 s) = (4.2 m/s) / (1.24 s) ≈ 3.4 m/s²

4 0

2 years ago

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0

2 years ago

Select the correct answer.

lutik1710 [3]

I’m sorry i haven’t found the answer to this

8 0

3 years ago

A spring-mass system has a spring constant k = 383 N/m, length of the spring L = 0.5 m, and the mass attached to it is M = 3.8 k

sergiy2304 [10]

Answer:

Frequency = f = 10.0394 (1/s)

Explanation:

The frequency of oscillation of the system is given by the action:

f= √(k/m)

f= system count

k = spring constant

m = mass connected to the spring

Therefore the frequency will be:

f= √(k/m) = √(383(N/m) / (3.8kg))= √( 100.7895 (kg×m/s²)/(kg ) =

= √( 100.7895 (1/s²) = 10.0394 (1/s)

4 0

3 years ago