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2 years ago

The heat of fusion for water is the amount of energy needed for water to

Physics

2 answers:

beks73 [17]2 years ago

6 0

The heat of fusion for water is the amount of energy needed for water to <span>melt. (c)</span>

omeli [17]2 years ago

5 0

<span>The heat of fusion for water is the amount of energy needed for water to
boil

</span>

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Why is it expensive to bring electricity into urban areas?

motikmotik

Because of the pole and the generator you would have to biuld

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3 years ago

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A coin purse contains d dimes and q quarter's. There are 20 coins in the purse and the total value of the coins is $4.25. Which

ch4aika [34]

Answer:

Explanation:

4 0

2 years ago

A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm chegg

Evgen [1.6K]

Fundamental frequency,

f=v2l=T/μ−−−−√2l

=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√

=58.96Hz

Let, n th harmonic is the hightest frequency, then

(58.93)n = 20000

∴N=339.38

Hence, 339 is the highest frequency.

∴fmax=(339)(58.93)Hz=19977Hz.

<h3>What is frequency?</h3>

In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.

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brainly.com/question/254161

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7 0

1 year ago

A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

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2 years ago

A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit

Kipish [7]

Answer:

F = 47.6 N

Explanation:

Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

$F = \frac{\Delta p}{\Delta t}$

So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:

$F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N$

⇒ Fsk = 47.6 N (normal to the wall)

3 0

3 years ago