Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.
Your answer is: B) Increasing the number of coils around the core
Have an amazing day and stay hopeful!
Explanation:
It is known that relation between torque and angular acceleration is as follows.
![\tau = I \times \alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I%20%5Ctimes%20%5Calpha)
and, I = ![\sum mr^{2}](https://tex.z-dn.net/?f=%5Csum%20mr%5E%7B2%7D)
So, ![I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}](https://tex.z-dn.net/?f=I_%7B1%7D%20%3D%202%20kg%20%5Ctimes%20%281%20m%29%5E%7B2%7D%20%2B%202%20kg%20%5Ctimes%20%281%20m%29%5E%7B2%7D)
= 4 ![kg m^{2}](https://tex.z-dn.net/?f=kg%20m%5E%7B2%7D)
![\tau_{1} = 4 kg m^{2} \times \alpha_{1}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%20%3D%204%20kg%20m%5E%7B2%7D%20%5Ctimes%20%5Calpha_%7B1%7D)
![\tau_{2} = I_{2} \alpha_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%20%3D%20I_%7B2%7D%20%5Calpha_%7B2%7D)
So, ![I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}](https://tex.z-dn.net/?f=I_%7B2%7D%20%3D%202%20kg%20%5Ctimes%20%280.5%20m%29%5E%7B2%7D%20%2B%202%20kg%20%5Ctimes%20%280.5%20m%29%5E%7B2%7D)
= 1 ![kg m^{2}](https://tex.z-dn.net/?f=kg%20m%5E%7B2%7D)
as ![\tau_{2} = I_{2} \alpha_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%20%3D%20I_%7B2%7D%20%5Calpha_%7B2%7D)
=
Hence, ![\tau_{1} = \tau_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%20%3D%20%5Ctau_%7B2%7D)
![4 \alpha_{1} = \alpha_{2}](https://tex.z-dn.net/?f=4%20%5Calpha_%7B1%7D%20%3D%20%5Calpha_%7B2%7D)
Thus, we can conclude that the new rotation is
times that of the first rotation rate.
Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
Answer:
I dont get what your asking sorry boo
Explanation: