Answer:
3 m/s
Explanation:
Average Speed = ![\frac{Total Distance}{Total time}](https://tex.z-dn.net/?f=%5Cfrac%7BTotal%20Distance%7D%7BTotal%20time%7D)
Plug in the numbers, it will be (6m + 3m) divided by (2s + 1s), which is 9m/3s, which equals to 3m/s.
Hey there!
Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.
0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s
–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s
So, the cars are traveling at -0.33 m/s in the direction of the second car.
Hope this helps
<em>Tobey</em>
Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s
Answer:
44.13015
Explanation:
use the 9.8067 newtons to 1 kg conversion
Answer:
68cm
Explanation:
You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get
![p_f=p_i\\mv_1+Mv_2=(m+M)v](https://tex.z-dn.net/?f=p_f%3Dp_i%5C%5Cmv_1%2BMv_2%3D%28m%2BM%29v)
m: mass of the bullet
M: mass of the pendulum
v1: velocity of the bullet = 410m/s
v2: velocity of the pendulum =0m/s
v: velocity of both bullet ad pendulum joint
By replacing you can find v:
![(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}](https://tex.z-dn.net/?f=%280.038kg%29%28410m%2Fs%29%2B0%3D%280.038kg%2B4.2kg%29v%5C%5C%5C%5Cv%3D3.67%5Cfrac%7Bm%7D%7Bs%7D)
this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:
![E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=E_%7Bfp%7D%3DE_%7Bki%7D%5C%5C%5C%5C%28m%2BM%29gh%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
g: 9.8/s^2
h: height
By doing h the subject of the equation and replacing you obtain:
![(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m](https://tex.z-dn.net/?f=%280.038kg%2B4.2kg%29%289.8m%2Fs%5E2%29h%3D%5Cfrac%7B1%7D%7B2%7D%280.038kg%2B4.2kg%29%283.67m%2Fs%29%5E2%5C%5C%5C%5Ch%3D0.68m)
hence, the heigth is 68cm