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Oksanka [162]
3 years ago
7

Science question: How do humans use the magnetic field for navigation?

Physics
1 answer:
I am Lyosha [343]3 years ago
4 0

Answer:

Magnetoreception (also magnetoception) is a sense which allows an organism to detect a magnetic field to perceive direction, altitude, or location. This sensory modality is used by a range of animals for orientation and navigation, and as a method for animals to develop regional maps.

Explanation:

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If you borrow $12250 and pay $147 in annual interest, the APR on your loan is 1.2% 14.7% 1.23% 7.35%
ziro4ka [17]

Answer: 1.2%

Explanation:

Given

If one borrows $12,250

and give $147 interest on it

Then the interest is given from the formula

\Rightarrow S.I=\dfrac{P\times R\times T}{100}\\\\\Rightarrow 147=\dfrac{12,250\times R\times 1}{100}\\\\\Rightarrow R=\dfrac{147}{122.50}\\\\\Rightarrow R=1.2\%

Thus, the annual rate of interest  is 1.2%

7 0
3 years ago
onsider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST applies
cluponka [151]
The answer is C) A girl hangs by both hands, motionless, from a trapeze.
5 0
3 years ago
A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m
Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0
3 years ago
A lamp hangs vertically from a cord in a descending elevator that decelerates at 1.7 m/s2. (a) if the tension in the cord is 63
zubka84 [21]

(a)
The formula is: 
∑ F = Weight + T = mass * acceleration 

as the elevator and lamp are moving downward, I choose downward forces to be positive. 
Weight is pulling down = +(9.8 * mass) 
Tension is pulling up, so T = -63 
Acceleration is upward = -1.7 m/s^2 

(9.8 * mass) + -63 = mass * -1.7 
Add +63 to both sides 
Add (mass * 1.7) to both sides 

(9.8 * mass) + (mass * 1.7) = 63 
11.5 * mass = 63

mass = 63 / 11.5 

Mass = 5.48 kg 


(b)
Since the elevator and lamp are going upward, I choose upward forces to be positive. 
Weight is pulling down = -(9.8 * 5.48) = -53.70 
Acceleration is upward, so acceleration = +1.7 


-53.70 + T = 5.48 * 1.7

T = 53.70 + 9.316 = approx 63 N 

The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated upward at the same rate of 1.7 m/s^2

3 0
3 years ago
An archer pulls back the string of a bow to release an arrow at a target. Which kind of potential energy is transformed to cause
Morgarella [4.7K]
It is elastic energy
8 0
4 years ago
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