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matrenka [14]
3 years ago
15

student built a simple heat engine that could lift masses. If the heat engine takes in 300 J of heat and loses 50 J to the envir

onment, how much height will it lift a 1.5 kg mass?
Physics
1 answer:
k0ka [10]3 years ago
4 0

Answer:

So it will lift the mass by h = 17 m

Explanation:

As per energy conservation we know that

Q_1 = W + Q_2

here we know that

Q_1 = 300 J

Q_2 = 50 J

now we have

W = 300 - 50

W = 250 J

so work done by the engine is 250 J

now we have

W = mgh

250 = 1.5 \times 9.81 \times h

h = 17 m

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Alex

Answer:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

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What is a materials ability to be dissolved in a solvent
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Answer: Solubility

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4 years ago
(No links, if you aren’t gonna answer here don’t answer at all.)
Liula [17]

Answer:

D. Atoms with eight valence electrons usually do not gain or lose

electrons.

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Based on the octet rule which states that atoms of chemical elements gain, lose or share electrons so as to have eight (8) electrons in their valence shell. Therefore, atoms of chemical elements bond in order to attain the electronic configuration of a noble gas i.e a full valence shell which comprises of eight (8) electrons.

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6 0
4 years ago
An object with mass of 4kg is thrown with initial velocity of 20m/s from point A and follows the track of ABCD.
soldier1979 [14.2K]

<u>The distance of the length YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach the ground is 0.17s</u>

Data given;

  • Mass = 4kg
  • Initial velocity = 20m/s
  • length CD = 5m (from the image given)

a)

<h3>Determine The Length of YD</h3>

YD=CDsinθ = 5sin45=\frac{5}{\sqrt{2} } = 3.57m

b)

<h3>The velocity of the object at point D</h3>

The change in kinetic energy is given as

Δ in kinetic energy = Δ in potential energy + work done by friction

K.E - 1/2mv^2 = mgh_1 - mgh_2 + (-μmg.x)

K.E = mg(50 - 3.57) + (-mg(0.3*100) + 1/2 mv^2

\frac{1}{2}mv^2_f=mg(46.43)-mg(30)+\frac{1}{2} m(400)\\\\v^2_f=20(16.43)+400\\v^2_f=728.6\\v=\sqrt{728.6} \\v_f=26.99 = 27m/s

The velocity of the object at D with a distance of 5m.

c)

<h3>The the required for the object to reach ground</h3>

The velocity of the object in the y-axis is

v_y=vsin45=19.09

Acceleration in y-axis = 9.8

Height = 3.57m

h = ut+\frac{1}{2}at^2

3.57=19.28(t)+\frac{1}{2}(9.8)t^2\\3.57=19.28t+4.9t^2\\4.9t^2+19.28t-3.57=0\\a=4.9, b=19.28, c= -3.57\\t=\frac{-19.28+\sqrt{(19.28)^2-4(4.9)(-3.57)} }{(2*4.9)} \\t=\frac{-19.28+21.02}{9.8}

Taking the positive value

\frac{-19.28+21.02}{9.8}=0.17s

The time required for the object to reach ground is 0.17s

From the calculations above,<u> the distance YD is 3.57m, the velocity of the object at point D is 27m/s and the time required for the object to reach ground is 0.17s</u>

Learn more on projectile motion here;

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5 0
3 years ago
Why is the international bureau of weight and measure established​
Reptile [31]

Answer:

Explanation:

International Bureau of Weights and Measures (BIPM), French Bureau International des Poids et Mesures, international organization founded to bring about the unification of measurement systems, to establish and preserve fundamental international standards and prototypes, to verify national standards, and to determine standard units.

6 0
3 years ago
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