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Readme [11.4K]
3 years ago
14

1- A train travels 92 kilometers in 3 hours, and then 55 kilometers in 3 hours. What is its average speed?

Physics
1 answer:
Hunter-Best [27]3 years ago
8 0
<h3>Answer:</h3>
  1. 24.5 km/h
  2. 4 17/27 m/s
  3. 11/3 m/s²
<h3>Explanation:</h3>

1. The average speed is the ratio of total distance to total time:

... speed = distance/time = (92 km +55 km)/(3 h +3h) = (147 km)/(6 h)

... = 24.5 km/h

2. speed = distance/time = (125 m)/(27 s) = 4 17/27 m/s

3. a = ∆v/∆t = (15 m/s -4 m/s)/(3 s) = 11/3 m/s²

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A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the
REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0
3 years ago
What is the wavelength of a light of frequency 7.21?
erik [133]
MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v  E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s  so you are left with Solution 4.78 10-19 J OR .478 aJ  <span>Apex - 467 nm ^.^  hopefully thats the correct thing</span>
4 0
3 years ago
Most of the Earth's volcanoes occur
timofeeve [1]

Answer:

A.

Explanation:

NEAR THE CENTER OF TECTONIC PLATES.

3 0
2 years ago
Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express
AlladinOne [14]

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

          Vb = k Q / r        r <R

         

We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

         Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0
3 years ago
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

   The  focal length of  B  is  f_{objective } =  43.0 \ cm

    The focal length of  A  is   f_{eye} =  10.4 \  cm

The  theoretical angular  magnification is mathematically represented as

           m = \frac{f_{objective }}{f_{eye}}  =  \frac{43.0}{10.4}

            m = \frac{f_{objective }}{f_{eye}}  =  4.175

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular  magnification lies within the angular magnification range

3 0
3 years ago
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