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quester [9]
3 years ago
9

If the latent heat of fusion of ice is 334kj/kg and the specific heat of water is 4.18kj kg-1 k-1 how much heat, in joules, must

6.5g of ice absorb to reach the final temperature in part a?
Physics
1 answer:
lord [1]3 years ago
4 0
The answer is 500 because it is 
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A 300-kg piano being held by a crane is accidentally dropped from a height of 15 meters. a. What is the speed of the piano just
FinnZ [79.3K]

Answer:

a) 17.16m/s

b) 44,145J

c) Sound the piano makes when hitting the ground, vibration of the ground, heat.

d) i) It's smaller due to the energy dissipated by the friction between air and the parachute.

ii) It stays the same, the only difference is that the dissipated energy is distributed between air resistance and the kinetic energy dissipated by the ground whent he piano hits it.

Explanation:

a)

In order to solve this problem we must start by doing a drawing of the situation, which will help us visualize the problem better. (See attached picture).

So, in this problem we can ignore air resistance so we can say that the energy is conserved, this is the total initial energy is the same as the total final energy, so we get that:

U_{0}+K_{0}=U_{f}+K_{f}

When the piano is released it has an initial speed of zero, so the initial kinetic energy is zero. When the piano hits the ground it will have a height of 0m, so the final potential energy is zero as well. This will simplify our equation:

U_{0}=K_{f}

We know that potential energy is given by the formula:

U=mgh

and kinetic energy is given by the formula:

K=\frac{1}{2}mv^{2}

which can be substituted in our equation:

mgh=\frac{1}{2}mv^{2}

we can divide both sides of the equation into the mass of the piano, so we get:

gh=\frac{1}{2}v^{2}

which can be solved for the final velocity which yields:

v=\sqrt{2gh}

we can now substitute the data provided by the problem so we get:

v=\sqrt{2(9.81m/s^{2})(15m)}

which yields:

v=17.16m/s

b)

Since energy is conserved, this means that the total dissipated energy will be the same as the potential energy, so we get that:

E=mgh

so

E=(300kg)(9.81m/s^{2})(15m)

which yields:

E=44,145J

c)

When the piano hits the ground, the kinetic energy it had will be transformed to other types of energy, mostly vibration and heat. The vibration will turn to sound due to the movement of air created by the piano itself and the ground. And heat is created by the friction between the molecules created by the vibrations and the collition itself. So some of the indicators of this release of energy could be:

-Sound

-Vibration

-Heat.

d)

i) The amount of inetic energy dissipated would decrease due to the friction between air and the parachute. Since air is resisting the movement of the piano, this will translate into a loss of energy, if we did an energy balance we would get that:

U_{0}=K_{f}+E_{p}

The total amount of energy is conserved but it will be distributed between the energy lost due to air resistance and the kinetic energy the piano has at the time it hits the ground.

ii) So the total amount of energy dissipated remains the same, the only difference is that it will be distributed between air resistance and the kinetic energy of the piano.

3 0
3 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
2 years ago
The passenger-side rear view mirror on a car says, “Objects in the mirror may be closer than they appear”. Assuming the images a
Ipatiy [6.2K]

Answer:

the mirror is a convex mirror

4 0
3 years ago
A 55 kilogram person jumps off a cliff and hits the water 5.8 seconds later, how high is the cliff above the water?
dusya [7]
The cliff is 319. kilograms above the water
5 0
3 years ago
Monochromatic light falling on two very narrow slits 0.048mm apart. Successive fringes on a screen 5.00m away are 6.5cm apart ne
tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}

<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>

Now, the frequency can be given as:

f = \frac{c}{\lambda}

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\

<u>f = 5.13 x 10¹⁴ Hz</u>

5 0
2 years ago
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