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zlopas [31]
2 years ago
6

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

complete. What is the magnitude of the centripetal acceleration during the turn?
a. 12 m/s2
b. 9.8 m/s2
c. 0 m/s2
d. 6.9 m/s
e. 8.1 m/s2
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

The magnitude of the centripetal acceleration during the turn is a=12.04\ m/s^2.

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , T=t\times 4=60\ s.

Now , Velocity ,

v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.

Also , we know :

Centripetal acceleration ,

a=\dfrac{v^2}{r}

Putting all values we get :

a=12.04\ m/s^2.

Hence , this is the required solution .

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Answer:

A ferromagnetic material is a temporary magnet. The domains in a ferromagnetic material are randomly arranged. Under certain actions, the domains align in a particular direction and the material acts as a magnet. The actions that can cause alignment of domains in a ferromagnetic material are:

  • rubbing the material against a magnet would cause the alignment of domains in the same direction as of the magnet.
  • passing electricity around the material would generate magnetic field which would cause domains to align along the direction of the field.
  • placing the material near a strong magnet would cause the alignment of domains in the direction of the field generated by the strong magnet.

Other actions like heating the material,  placing the material in a magnetic field of opposite polarity and hitting the material would lead to demagnetization of the magnetic material.

8 0
2 years ago
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The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.
Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

W=0.24\times 1\times 10^3=240\ W/m^2

Energy required by the United States is 1\times 10^{19}\ J/yr (assumed)

Power

P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W

Area

A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

1\ m^2=\frac{1}{1000\times 1000}\ km^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2

Converting to mi²

1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0
2 years ago
A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha
Vlada [557]
The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
\Delta U = q \Delta V = q (V_f - V_i)
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: q=1.6 \cdot 10^{-19} C, and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J
7 0
2 years ago
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A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the
Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

      F = m a = m dv / dt

      dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

       dv = ∫ (3 + 2t) dt / m

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Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit   t = t and v = 0

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Let's look for the mass

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      m = W / g

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      m = 0.625 slug

Let's solve the second degree equation

     t² + 3t -3.75 = 0

     t = (-3 ± √ (32 + 4 1 3.75)) / 2

     t = (-3 ± 4,899) / 2

     t₁ = 0.95 s

     t₂ = -3.95 s

We take the positive time

6 0
2 years ago
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