Answer:
The circumference will be "43,200 km".
Explanation:
As you should be directly south of Tyene, this same angular distance among both yourself as well as the town of Tyene seems to be the distinction to be made throughout the height including its sun at two positions on a specified day.
The sun is therefore immediately overhead in Tyene, does have an altitude with respect to the horizon will be:
⇒ ![\Theta_r = 90^{\circ}](https://tex.z-dn.net/?f=%5CTheta_r%20%3D%2090%5E%7B%5Ccirc%7D)
At the very same day on yourself location, the altitude of the sun will be:
⇒ ![\Theta_A=85^{\circ}](https://tex.z-dn.net/?f=%5CTheta_A%3D85%5E%7B%5Ccirc%7D)
Therefore, a complete circle is 360 degrees. So the angular distinction will be somewhere between you as well as Tyene:
⇒ ![\frac{5}{360}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B360%7D)
Therefore,
⇒ ![\frac{\Theta}{x}=\frac{360}{Circumference}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CTheta%7D%7Bx%7D%3D%5Cfrac%7B360%7D%7BCircumference%7D)
⇒ ![Circumference=\frac{360}{\Theta}\times x](https://tex.z-dn.net/?f=Circumference%3D%5Cfrac%7B360%7D%7B%5CTheta%7D%5Ctimes%20x)
If, ![\Theta = 5^{\circ}](https://tex.z-dn.net/?f=%5CTheta%20%3D%205%5E%7B%5Ccirc%7D)
![x (separation) = 600 \ km](https://tex.z-dn.net/?f=x%20%28separation%29%20%3D%20600%20%5C%20km)
Now,
The circumference = ![\frac{360}{5}\times 600](https://tex.z-dn.net/?f=%5Cfrac%7B360%7D%7B5%7D%5Ctimes%20600)
= ![43,200 \ km](https://tex.z-dn.net/?f=43%2C200%20%5C%20km)
Answer:
a_x(3.98) = 7.0844 m/s^2
Explanation:
Given:
- The eastward component of car's velocity is given as follows:
v_x (t) = 0.890*t^2
For, 0 s < t < 5 s
Find:
What is the acceleration of the car when v_x = 14.1 m/s ?
Solution:
- We will first compute the time t at which the velocity v_x component equals 14.1 m /s :
14.1 = 0.890*t^2
t = sqrt (15.8427)
t = 3.98 s
- Now use the relation of v_x and derivative to obtain a_x acceleration in eastward direction:
a_x (t) = dv_x / dt
a_x(t) = 1.78*t
- Evaluate at t = 3.98 s:
a_x(3.98) = 1.783.98
a_x(3.98) = 7.0844 m/s^2
The electron’s speed when it reaches the positive plate is 2.65 X 10⁶m/s
<u>Explanation:</u>
Electric field strength, E = 20,000 N/C
Distance, x = 1 mm = 1 X 10⁻³m
Speed of the electron, s = ?
We know,
Charge of an electron, q = 1.602 X 10⁻¹⁹ C
So the force, F is
20000 N/C x 1.602 X 10⁻¹⁹ C = 3.2 X 10⁻¹⁵ Newtons
mass of electron is 9.1×10⁻³¹ kg
F = ma
a = 3.2 X 10⁻¹⁵ / 9.1 X 10⁻³¹ = 3.52 X 10¹⁵ m/s²
v = √(2ad) = √(2*3.52 X 10¹⁵ X 0.001) = 2.65 X 10⁶m/s
Therefore, the electron’s speed when it reaches the positive plate is 2.65 X 10⁶m/s