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2 years ago

9

Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o

n the hamburger.

2 answers:

egoroff_w [7]2 years ago

5 0

Sample Response: The piece of cheese melts when it’s placed on the hamburger because conduction occurs. The hamburger is warmer than the cheese, so thermal energy is transferred from the hamburger to the cheese, causing it to melt.

Correct one for e2020

Svetllana [295]2 years ago

4 0

If Kai takes the burger off of the grill and puts cheese on it, it melts because the burger still is hot from being on the grill. The heat in and on the burger doesn’t go away immediately, so that is how and why the cheese melts.

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A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

6 0

2 years ago

A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

= 3/2×(5.3)×(8.31451)×(24 + 273)

= 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0

2 years ago

How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236

Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(c) 7.03 = 3

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0

2 years ago

Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in centim

Andrej [43]

Answer:Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s

Explanation:

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2 years ago

You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming w

Elina [12.6K]

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

$v_f$ = Final velocity

$v_i$ = Initial velocity

$t_f$ = Final time

$t_i$ = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

$\Delta P=F\Delta t$

Force

$F=mg$

So,

$m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s$

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

$K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m$

The maximum height above your hand reached by the ball is 5.93505 m

5 0

2 years ago