Answer:
The car will travel 30 miles during the 30-minutes period of acceleration.
Explanation:
Given data :
Initial velocity = v₁ = 50 miles/hour
Final velocity = v₂ = 70 miles/hour
Time = t = 30 min = 0.5 hour
Using the definition of acceleration, we find the acceleration (a)
a = (v₂ - v₁) ÷ t
a = (70 - 50) ÷ 0.5
a = 20 ÷ 0.5
a = 40 miles/hour²
Using 3rd equation of motion, we find the distance travel (s)
2as = v₂² - v₁²
2(40)s = 70² - 50²
80 × s = 4900 - 2500
s = 2400 ÷ 80
s = 30 miles
Answer:
A radio telescope helped the astronomers discover the CMB.
Explanation:
- Penzias and Wilson while experimenting with a radio telescope in 1964, accidentally discovered the radiation that exists universally also known as the CMB.
- This was used to support the "Big Bang Theory" and not the "Steady State Theory"
- CMB is the faint cosmic radiation that fills up the universe. It provides important data for understanding early universe.
- This data tells us about the composition of the universe and its age which raises new questions about the universe.
Answer:
1. Electromagnetic waves travel in a vacuum whereas mechanical waves do not.
2. The ripples made in a pool of water after a stone is thrown in the middle are an example of mechanical wave. Examples of electromagnetic waves include light and radio signals.
3. Mechanical waves are caused by wave amplitude and not by frequency. Electromagnetic Waves are produced by vibration of the charged particles.
4. While an electromagnetic wave is called just a disturbance, a mechanical wave is considered a periodic disturbance.
Explanation:
The x-acis of a trajectory represents its C
Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
![\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%2Bmg%3D%5Cfrac%7Bmv%5E2%7D%7BL%7D_%7B%7D%20%5C%5C%20v%5E2%3D%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%7D%20%5Cend%7Bgathered%7D)
For the maximum velocity of the ball at the top of the vertical circular motion,
![v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T_%7B%5Cmax%20%7D%2Bmg%29%7D%7Bm%7D%7D)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%287.5_%7B%7D%2B0.31%5Ctimes9.8%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%2810.538%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B17.34%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D4.16%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
