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Hitman42 [59]
3 years ago
13

A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m

. When the object is a distance 1.25×10−2 m from its equilibrium position, it is observed to have a speed of 0.305 m/s . Part A What is the total energy of the object at any point of its motion
Physics
1 answer:
zzz [600]3 years ago
4 0

Answer:

310.38\times 10^{-4}j

Explanation:

We have given mass m=0.155 kg

Force constant K = 305 N/m

Distance X=1.25\times 10^{-2}m

Velocity v=.305 m/sec

The total energy at any position of the motion is give by E=\frac{1}{2}mv^2+KX^2  here \frac{1}{2}mv^2 is energy due to motion and KX^2  is energy due to spring elongation

So total energy E=\frac{1}{2}\times 0.155\times 0.305^2+305\times 0.0125^2=310.38\times 10^{-4}j

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weight = 4.6 x 9.81

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Which will result in positive buoyancy and cause the object to float?
Wewaii [24]

the answer should be:

When the buoyant force is equal to the force of gravity

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(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball
Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

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6 0
3 years ago
A rocket ship starts from rest and turns on its forward booster rockets causing it to have a constant acceleration of 4 meters p
bagirrra123 [75]

Complete question is;

A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?

Answer:

v = 12 m/s

Explanation:

We are given;

Initial velocity; u = 0 m/s (because ship starts from rest)

Acceleration; a = 4 m/s²

Time; t = 3 s

To find velocity after 3 s, we will use Newton's first equation of motion;

v = u + at

v = 0 + (4 × 3)

v = 12 m/s

6 0
3 years ago
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