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musickatia [10]
3 years ago
9

Heating a metal from room temperature to pouring temperature in a casting operation depends on all of the following properties e

xcept which one: (a) density, (b) heat of fusion, (c) melting temperature, (d) specific heat, or (e) thermal expansion?
Physics
1 answer:
lapo4ka [179]3 years ago
7 0

Answer:

(e) thermal expansion

Explanation:

The density, the heat of fusion and the melting temperature of the metal are determining characteristics to take into account to raise the temperature of the metal from room temperature to the melting temperature. Since they will determine the following:

Density: is the relationship between the mass of a body and the volume it occupies in outer space.

Heat of fusion: The enthalpy of fusion or heat of fusion is the amount of energy needed to make a mole of an element that is at its melting point pass from the solid state to the liquid, at constant pressure.

Melting temperature is defined as the temperature at which the phase transition from the solid state to the liquid occurs at normal atmospheric pressure.

While the dilution of metals will only have an influence on the volume it will occupy but not on the heating process

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

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

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

4 0
3 years ago
Read 2 more answers
At what speed, as a fraction of c , is a particle's total energy twice its rest energy
WINSTONCH [101]
The rest energy of a particle is
E_0=m_0 c^2
where m_0 is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
E=mc^2 =  \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
E=2E_0
which means:
\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2
\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2
From which we find the ratio between the speed of the particle v and the speed of light c:
\frac{v}{c}=  \sqrt{1- (\frac{1}{2})^2 }  =0.87
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
3 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
Please can someone answer this question ​
EastWind [94]
B????????? I thinkkkkk
7 0
3 years ago
Read 2 more answers
If a thermometer measured the temperature in an oven as 400oF five days in a row when the temperature was actually 397oF, this m
aleksandrvk [35]

Answer:

It can be said to be reliable although it is not valid

Explanation:

This is because Reliability means an indicator of consistency, A measure should produce similar or the same results consistently if it measures the same quantity. So does the thermometer measures over 5days but it is not valid because it deviates from the real value

5 0
3 years ago
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