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Aleksandr-060686 [28]
3 years ago
13

What are the methods of heat transfer? ​

Physics
1 answer:
m_a_m_a [10]3 years ago
8 0

Answer:

Radiation , Conduction and Convection

Explanation:

Those are the ways heat is transferred

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The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscop
Blababa [14]

Answer:

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

   =18/2.76

fₙ =  6.52 cm

The focal length of eye piece is 6.52 cm.

6 0
3 years ago
2. Find the velocity when the displacement is 40 m and the time is 5 s.
LenKa [72]

Answer:

V=?

S= 40m

t=5s

V=S/t

V=40/5

V=8m/s

3 0
3 years ago
Two charges separated by one meter exert a 9 N force on each other. If the charges are pushed to a 3 meter separation, the force
tamaranim1 [39]

Answer:

False

Explanation:

The formula of force that exists between two charges is expressed as;

F = kq1q2/r²

If two charges separated by one meter exert a 9 N force on each other, the;

9 = kq1q2/1²

9 = kq1q2 ..... 1

If the charges are pushed to a 3 meter separation, then;

F =  kq1q2/3²

F =  kq1q2/9 .... 2

Divide both equations;

9/F = (kq1q2)/ kq1q2/9

9/F =  kq1q2 * 9/ kq1q2

9/F = 9

F = 9/9

F = 1N

Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False

3 0
3 years ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft
erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)

Now, differentiating both sides with respect to time 't', we get:

2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}

Now, we are given speed as 25 mph. So, \frac{de}{dt}=25\ mph

Also, when e=91\ mi, we can find 'x' using equation (1). This gives,

x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi

Now, plug in the values of 'e' and 'x' and solve for \frac{dx}{dt}. This gives,

\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0
3 years ago
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