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3 years ago

What happens when the speed of an object increase without change in height in the potential energy?

1 answer:

7 0

You're describing an increase in the object's kinetic energy, which depends
on speed, and no change in its potential energy, which depends on height.

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As light shines from air to another medium, i = 42.0Â°. The light bends toward the normal and refracts at 38.0° What is

eduard

Answer:

1.09

Explanation:

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3 years ago

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea

grandymaker [24]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

$A = 1.803 ft$

Period = $\frac{\pi}{2}$ seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = $\frac{w}{g}$

m= $\frac{32}{32}$

= 1 slug

Also from Hooke's Law

2 k = 32

k = $\frac{32}{2}$

k = 16 lb/ft

Using the function:

$\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0$

$x(0) = -1$ (because of the initial position being above the equilibrium position)

$x(0) = -6$ ( as a result of upward velocity)

NOW, we have:

$x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)$

However;

$x(0) = -1$ means

$-1 =c_1\\c_1 = -1$

$x(0) =-6$ also implies that:

$-6 =4(c_2)\\c_2 = - \frac{6}{4}$

$c_2 = -\frac{3}{2}$

Hence, $x(t) =-cos4t-\frac{3}{2} sin 4t$

$A = \sqrt{C_1^2+C_2^2}$

$A = \sqrt{(-1)^2+(\frac{3}{2})^2 }$

$A=\sqrt{\frac{13}{4} }$

$A= \frac{1}{2}\sqrt{13}$

$A = 1.803 ft$

Period can be calculated as follows:

= $\frac{2 \pi}{4}$

= $\frac{\pi}{2}$ seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

$x* \frac{\pi}{2} = 4 \pi$

$x \pi = 8 \pi$

$x=8$ cycles

5 0

3 years ago

A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin

blsea [12.9K]

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

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2 years ago

Susan gently pushes the tip of her finger against the eraser on her pencil and the pencil does not move. Which of the

pochemuha

Answer: D. ➡️⬅️

Explanation: I just knew the answer ;)

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2 years ago

Plz help me! <br><br> How do electromagnetic fields work?

MrMuchimi

Answer:

Electromagnetic field, a property of space caused by the motion of an electric charge. A stationary charge will produce only an electric field in the surrounding space. If the charge is moving, a magnetic field is also produced. An electric field can be produced also by a changing magnetic field.

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3 years ago