Chunking helps students identify key words and ideas, develops their ability to paraphrase, and makes it easier for them to organize and synthesize information. Please like
Answer:
- 210 rad/s²
Explanation:
n = frequency of rotation = 3400/60 = 170/3 per sec.
angular velocity ω ( 0 ) at time 0 = 2π n = 2π x 170/3
angular velocity at time t = ω(t) = 0
now, ω²( t) = w²(o) + 2α Φ ( α = angular acceleration and Φ = angular displacement) = 2π x 48 rad.
0 = ( 2π x 170/3 )² + 2α x 48 x 2π
α = - (2π x 170 x 170 )/ (3 x 3 x 2 x 48 ) = 210 rad / s²
Answer:
We cannot tell from the information given
Explanation:
Given;
mass of the box, m = 5 kg
first force, F₁ = 10 N
second force, F₂ = 5 N
(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;
∑Fx = 10 N - 5 N
= 5 N
Apply Newton's second law of motion;
∑Fx = ma
a = ∑Fx/m
a = 5 / 5
a = 1 m/s² in the direction of the 10 N force.
(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;
∑Fx = 10 N + 5 N
∑Fx = 15 N
a = 15 / 5
a = 3 m/s²
Therefore, the information given is not enough to determine the acceleration of the box.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Explanation:
The total momentum before and after the collision must be conserved.
The total momentum before the collision is:
![p_i = m_1 v_1 + m_2 v_2](https://tex.z-dn.net/?f=p_i%20%3D%20m_1%20v_1%20%2B%20m_2%20v_2)
where m1 and m2 are the masses of the two players, and
![v_1](https://tex.z-dn.net/?f=v_1%20)
and
![v_2](https://tex.z-dn.net/?f=v_2)
their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.
The final momentum is instead
![p_f = (m_1+m_2)v_f](https://tex.z-dn.net/?f=p_f%20%3D%20%28m_1%2Bm_2%29v_f)
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.
By requiring that the momentum is conserved
![p_i=p_f](https://tex.z-dn.net/?f=p_i%3Dp_f)
we can calculate vf, the post-collision speed:
![m_1 v_1 + m_2 v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_1%20v_1%20%2B%20m_2%20v_2%20%3D%20%28m_1%2Bm_2%29v_f)
![v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 +m_2}= \frac{(98 kg)(8.6 m/s)+(76 kg)(9.8m/s)}{98 kg+76 kg}=9.1 m/s](https://tex.z-dn.net/?f=v_f%20%3D%20%20%5Cfrac%7Bm_1%20v_1%20%2B%20m_2%20v_2%7D%7Bm_1%20%2Bm_2%7D%3D%20%5Cfrac%7B%2898%20kg%29%288.6%20m%2Fs%29%2B%2876%20kg%29%289.8m%2Fs%29%7D%7B98%20kg%2B76%20kg%7D%3D9.1%20m%2Fs%20%20)
and the direction is the same as the direction of the players before the collision.