Answer:
The kinetic energy is 16 joules.
Explanation:
Kinetic energy (E) is given by:

Where:
m: is the mass
v: is the speed
When the kinetic energy is 4 joules, the speed of the object is:
Now, if the speed is increased twice then we have:
![E_{2} = \frac{1}{2}mv_{2}^{2} = \frac{1}{2}m(2v_{1})^{2} = \frac{1}{2}m[4(\frac{8}{m})] = 16 J](https://tex.z-dn.net/?f=%20E_%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv_%7B2%7D%5E%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm%282v_%7B1%7D%29%5E%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm%5B4%28%5Cfrac%7B8%7D%7Bm%7D%29%5D%20%3D%2016%20J%20)
Therefore, the kinetic energy is 4 times the initial value.
I hope it helps you!
Answer:
- 8.33 x 10⁻³ rad /s ( anticlockwise)
Explanation:
The rotational movement of beetle and turntable is caused by torque generated by internal forces , we can apply conservation of angular momentum.
That is ,
I₁ ω₁ = I₂ω₂ , ω₁ and ω₂ are angular velocity of beetle and turntable respectively.
ω₁ + ω₂ = .05 radian /s ( given )
Momentum of inertia of beetle I₁ = mass x (distance from axis)²
= 15 x 10⁻³ x R² ( R is radius of the turntable )
Momentum of inertia of turntable I₂ =1/2 mass x (distance from axis)²
= 75/2 x 10⁻³ x R² ( R is radius of the turntable )
I₁ ω₁ = I₂ω₂ ,
15 x 10⁻³ x R² x ( .05 - ω₂ ) = 75/2 x 10⁻³ x R² ω₂
15 x ( .05 - ω₂ ) = 75/2 x ω₂
.75 - 15ω₂ = 37.5ω₂
.75 = 52.5 ω₂
ω₂ = - 14.3 x 10⁻³ rad /s ( anticlockwise)
Answer:
a)Current will flow perpendicularly.
b)Magnitude of flux will be 2.987 N m2 C−1
Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?
B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?
Answer:
The frequency change percentage is 9.94%
Explanation:
The frequency shift can be calculated as follows.

= 
=9.95 GHz
So the frequency change seen by the detector is 9.95 - 9.05
% difference
= 9.94%