Answer:
Part a)
![t = 3.85 s](https://tex.z-dn.net/?f=t%20%3D%203.85%20s)
Part b)
![h = 72.67 m](https://tex.z-dn.net/?f=h%20%3D%2072.67%20m)
Part C)
![v_x = 25.98 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2025.98%20m%2Fs)
![v_y = 0](https://tex.z-dn.net/?f=v_y%20%3D%200)
Part d)
In horizontal direction velocity will remain constant
![v_x = 30 cos30 = 25.98 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2030%20cos30%20%3D%2025.98%20m%2Fs)
in vertical direction we have
![v_y = -22.77 m/s](https://tex.z-dn.net/?f=v_y%20%3D%20-22.77%20m%2Fs)
Explanation:
Part a)
Horizontal speed of the cannon
![v = 30.0 m/s](https://tex.z-dn.net/?f=v%20%3D%2030.0%20m%2Fs)
angle of projection
![\theta = 30^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2030%5Eo)
now we have
horizontal speed = ![v_x = vcos30 = 30 cos30 =25.98 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20vcos30%20%3D%2030%20cos30%20%3D25.98%20m%2Fs)
vertical speed = ![v_y = vsin30 = 30 sin30 = 15 m/s](https://tex.z-dn.net/?f=v_y%20%3D%20vsin30%20%3D%2030%20sin30%20%3D%2015%20m%2Fs)
now the time taken by it to cover the distance 100 m from the wall
![x = v_x t](https://tex.z-dn.net/?f=x%20%3D%20v_x%20t)
![100 = 25.98 t](https://tex.z-dn.net/?f=100%20%3D%2025.98%20t)
![t = 3.85 s](https://tex.z-dn.net/?f=t%20%3D%203.85%20s)
Part b)
Since it hits the ground in the same time
so the height of the castle is given as
![h = \frac{1}{2}gt^2](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![h = \frac{1}{2}(9.81)(3.85^2)](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B1%7D%7B2%7D%289.81%29%283.85%5E2%29)
![h = 72.67 m](https://tex.z-dn.net/?f=h%20%3D%2072.67%20m)
Part C)
At highest point of the projection
the vertical component of the velocity will become zero
so we will have
![v_x = 25.98 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2025.98%20m%2Fs)
![v_y = 0](https://tex.z-dn.net/?f=v_y%20%3D%200)
Part d)
In horizontal direction velocity will remain constant
so we have
![v_x = 30 cos30 = 25.98 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2030%20cos30%20%3D%2025.98%20m%2Fs)
in vertical direction we have
![v_y = v_i + at](https://tex.z-dn.net/?f=v_y%20%3D%20v_i%20%2B%20at)
![v_y = 15 - 9.81(3.85)](https://tex.z-dn.net/?f=v_y%20%3D%2015%20-%209.81%283.85%29)
![v_y = -22.77 m/s](https://tex.z-dn.net/?f=v_y%20%3D%20-22.77%20m%2Fs)
Part e)
The answer is B. A and C. Spring tides occur during a new moon and full moon
Answer:
d = 3.54 x 10⁴ Km
Explanation:
Given,
The distance between the two objects, r = 2.5 x 10⁴ Km
The gravitational force between them, F = 580 N
The gravitational force between the two objects is given by the formula
F = GMm/r² newton
When the gravitational force becomes half, then the distance between them becomes
Let us multiply the above equation by 1/2 on both sides
( 1/2) F = (1/2) GMm/r²
= GMm/2r²
= GMm/(√2r)²
Therefore, the distance becomes √2d, when the gravitational force between them becomes half
d = √2r = √2 x 2.5 x 10⁴ Km
= 3.54 x 10⁴ Km
Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.
The object with the larger radius has a faster tangential speed. Tangential speed is related to both rotational speed and radial distance from the rotating axis.
<h3>What is uniform circular motion?</h3>
Uniform circular motion is a type of motion of a particle around a circle at a constant speed. The magnitude of the speed of the particle is constant.While the direction is changing continuously.
Tangential speed is related to both rotational speed and radial distance from the rotating axis.
The object with the larger radius has a faster tangential speed.
Hence, option C is correct.
To learn more about the uniform circular motion, refer to the link;
brainly.com/question/2285236
#SPJ1
Answer:
![c>d>f=a>b>e](https://tex.z-dn.net/?f=c%3Ed%3Ef%3Da%3Eb%3Ee)
Explanation:
When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.
According to the Snell's law:
![\rm refractive\ index\ (n)=\frac{speed\ of\ light\ in\ the\ incident\ medium}{speed\ of\ light\ in\ the\ refracted\ medium}](https://tex.z-dn.net/?f=%5Crm%20refractive%5C%20index%5C%20%28n%29%3D%5Cfrac%7Bspeed%5C%20of%5C%20light%5C%20in%5C%20the%5C%20incident%5C%20medium%7D%7Bspeed%5C%20of%5C%20light%5C%20in%5C%20the%5C%20refracted%5C%20medium%7D)
a)
![n_1-n_2=1.33-1.00\\=0.33](https://tex.z-dn.net/?f=n_1-n_2%3D1.33-1.00%5C%5C%3D0.33)
b)
![n_2-n_1=1.46-1.33](https://tex.z-dn.net/?f=n_2-n_1%3D1.46-1.33)
![=0.23](https://tex.z-dn.net/?f=%3D0.23)
c)
![n_2-n_1=2.42-1.33\\=1.09](https://tex.z-dn.net/?f=n_2-n_1%3D2.42-1.33%5C%5C%3D1.09)
d)
![n_2-n_1=1.46-1.00\\=0.46](https://tex.z-dn.net/?f=n_2-n_1%3D1.46-1.00%5C%5C%3D0.46)
e)
![n_1-n_2=1.50-1.33\\=0.17](https://tex.z-dn.net/?f=n_1-n_2%3D1.50-1.33%5C%5C%3D0.17)
f)
![n_2-n_1=1.33-1.00\\=0.33](https://tex.z-dn.net/?f=n_2-n_1%3D1.33-1.00%5C%5C%3D0.33)
![c>d>f=a>b>e](https://tex.z-dn.net/?f=c%3Ed%3Ef%3Da%3Eb%3Ee)