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Anit [1.1K]
3 years ago
10

Blood ki kami se konsa rog hota hai?​

Physics
2 answers:
Darya [45]3 years ago
5 0

Answer:

anaemia, low blood pressure etc.

Sergeeva-Olga [200]3 years ago
5 0

Answer:

dnvfkwvnwk vkfw njnjejgkjaffrvnmio3ncgu

Explanation:

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An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2
morpeh [17]

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

brainly.com/question/14244057

#SPJ

5 0
2 years ago
You are pushing a rock along level ground and making the rock speed up. How does the size of the force you exert on the rock com
Aneli [31]

Answer:

The forces are of the same magnitude just opposite directions

Explanation:

Newton's Third law tells us that if body A exerts a force on body B, then B exerts an equal but opposite force on body B. so as you continue to push the rock causing it to accelerate, the rock also continues to push back at you by an equal force in the opposite direction.

4 0
3 years ago
If velocity is in m/s and time is in sec, what is the unit of acceleration?​
WINSTONCH [101]
The unit of acceleration would be m/s² :)
6 0
2 years ago
In what frame of reference would you be at rest while riding in a car?
Alexeev081 [22]

In the frame of reference of anybody in the car.

8 0
3 years ago
Read 2 more answers
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
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