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2 years ago

Does solid give out or take in energy when it melts?

Physics

1 answer:

vampirchik [111]2 years ago

8 0

Explanation:

i think it give out energy when it melts

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Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce

Stells [14]

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

$\omega=\frac{v}{r}$

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

$|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}$

the maximum relative velocity is 0.04m/s

hope this helps!!

6 0

3 years ago

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if you dropped a ball while standing on the surface of mars at what rate would it accelerate toward the ground

Inessa05 [86]

First of all you need to have in mind the following data:
<span>Mass of Mars: 6.43 x 10^23 kg
Radius of Mars: 3.40 x 10^6 m
Formulas: F = G(m1)(m2)/(r^2), m1 = F(r^2)/G(m2), m2 = F(r^2)/G(m1), F = ma G = 6.67 x 10^-11
</span><span> We can say that the first and second objects with mass can be called as following:
m1= centre of Mars
m2 = will be the ball.
</span>The distance between them = the radious of mars.

8 0

3 years ago

Read 2 more answers

2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an

Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

$\delta = \frac{Pl}{AE}$

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

$\delta = \frac{Pl}{AE}$

$\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}$

$\delta = 0.1973mm$

Therefore the elongaton of the rod in a 200mm gage length is $0.1973mm$

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

$\upsilon = \frac{l_{a}}{l_{s}}$

Where,

$\upsilon =$ Poission's ratio

$l_{a}$ = Lateral strain

$l_{s}$ = Linear strain

$\upsilon = \frac{\delta/d}{\delta/l}$

$0.3 = \frac{\delta/22}{0.1973/200}$

$0.3(\frac{0.1973}{200}) = \frac{\delta}{22}$

$\delta = 6.5109*10^{-3}mm$

Therefore the change in diameter of the rod is $6.5109*10^{-3}mm$

5 0

3 years ago

Label these parts on the wave below: Amplitude, Wavelength, Crest, Trough, Rest Position

Leviafan [203]

Answer:

Wavelength is the distance between from one crest to another crest or from one trough to another trough. The amplitude is the distance from the midpoint to the crest or trough. Crest is the highest point of the or a wave. Tough is the lowest point of the or a wave. Rest position is the position where it lies on the midpoint line.

Explanation:

I need a diagram to label these parts.

5 0

3 years ago

How does moon changes position every day?

photoshop1234 [79]

It moves relative to the stars

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4 years ago