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2 years ago

Does a stove have mechanical energy

1 answer:

6 0

-- Any object has gravitational potential energy relative to any place
lower than where the object is. The stove in the kitchen has potential
energy relative to the basement floor.

-- If an object is not moving, then it has no kinetic energy. The stove has
no kinetic energy unless you throw it or drop it out of a window.

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A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te

MrMuchimi

<span>The force will be zero if the wagons are moving at a constant speed (i.e. not accelerating), as there is no frictional force to overcome. If the wagons are accelerating, the force will be proportional to the acceleration, and 20% of the force applied by A.</span>

5 0

2 years ago

(2) Put 5kg mass at left side (at 2m). This is fixed throughout the experiment! (3) Try to balance by putting 5kg mass at the ri

dmitriy555 [2]

We can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.

<h3>What is torque</h3>

The torque experienced by an object a given position is the product of the applied force and the perpendicular distance of the object.

When 5 kg mass is at 2 m on the left, another 5 kg at 2 m on the right will balance it.

$\tau _{net} = (2 \times 5 \times 9.8) - (2 \times 5 \times 9.8)\\\\\tau _{net} = 0$

<h3>Position of 10 kg mass on the right</h3>

Apply principle of moment

$F_1r_1 = F_2r_2\\\\(m_1gr_1) = (m_2gr_2)\\\\r_2 = \frac{m_1gr_1}{m_2g} \\\\r_2 = \frac{m_1 r_1}{m_2} \\\\r _2 = \frac{5 \times 2}{10} \\\\r_2 = 1 \ m$

<h3>Net torque</h3>

$\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (10 \times 9.8 \times 1) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0$

<h3>Position of the 20 kg mass</h3>

$r_2 = \frac{5 \times 2}{20} \\\\r_2 = 0.5 \ m$

<h3>Net torque</h3>

$\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (20 \times 9.8 \times 0.5) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0$

Thus, we can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.

Learn more about principles of moment here: brainly.com/question/26117248

7 0

1 year ago

Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?

Lelu [443]

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

5 0

2 years ago

A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

= 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A) angular frequency = ?

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k x - ω t)

y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)

4 0

2 years ago

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

2 years ago