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Flauer [41]
3 years ago
8

Heat is added to a substance, but its temperature does not rise. Which one of the following statements provides the best explana

tion for this observation?
Physics
2 answers:
GalinKa [24]3 years ago
7 0

Answer:

Heat is added to a substance, but its temperature does not rise. Which one of the following statements provides the best explanation for this observation? the substance must be a gas. the substance must be a non-perfect solid. the substance undergoes a change of phase. the substance has unusual thermal properties. the substance must be cooler than its environment.

hoa [83]3 years ago
6 0

Answer:

(A) The substance must be a gas.

B) The substance must be a non-perfect solid.

C) The substance undergoes a change of phase.

D) The substance has unusual thermal properties.

E) The substance must be cooler than its environment.

Answer is c was correct answer

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Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li
tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?​

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

                 W=9.81x  Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: (F_{J} +F_{T} =9.81x n)

F_{T} (510)=9.81x  (210+21.5)

F_{T} (510)=9.81 x (231.5)

F_{T} =4.452x N.

F_{J} =9.81x-4.452x

F_{J} =5.358 xN

Therefore F_{J} \geq F_{T}

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

#SPJ9

4 0
1 year ago
In your own words, explain how solar heating works
mestny [16]

Answer:

active solar heating systems use solar energy to heat a fluid either liquid or air and then transfer the solar heat directly to the interior space or to a storage system for later use. If the solar system cannot provide adequate space heating, an auxiliary or back-up system provides the additional heat.

hope this helps : )

4 0
3 years ago
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During an early phase of the planet's existence, earth's atmosphere was similar in composition to the atmospheres of mars and ve
KengaRu [80]

The answer is carbon dioxide. This primordial earths’ atmosphere was composed by gasses from degassing of the earth's interior after its formation. It is after the beginning of life that oxygen levels began to rise and levels of carbon dioxide began to reduce in the atmosphere (as a result of photosynthesis).

8 0
3 years ago
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)
cupoosta [38]

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b)  x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant  [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

2)

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

5 0
2 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
Read 2 more answers
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