Answer:
E. Student 1 is correct, because as θ is increased, h is the same.
Explanation:
Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.
Mathematically:
![F=m.g](https://tex.z-dn.net/?f=F%3Dm.g)
As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.
And for work we have:
![W=F.s\cos\theta](https://tex.z-dn.net/?f=W%3DF.s%5Ccos%5Ctheta)
where:
displacement of the object
angle between the force and displacement vectors
Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.
- In case of free-fall the angle between the force is and the displacement is zero.
- In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.
So, the work done by the gravitational force is same in the two cases.
Answer:
If you push horizontally with a small force, static friction establishes an equal and opposite force that keeps the book at rest. As you push harder, the static friction force increases to match the force. Eventually maximum static friction force is exceeded and the book moves.
Explanation:
Answer
given,
before collision
mass of car A = m_a = 1300 kg
velocity of car A = v_a = 35 mph
mass of car B = m_b= 1000 kg
velocity of car B = v_b = 25 mph
after collision
V_a = 30 mph
V_b = 31.5 mph
Initial momentum
![P_1 = m_av_a + m_b v_b](https://tex.z-dn.net/?f=P_1%20%3D%20m_av_a%20%2B%20m_b%20v_b)
![P_1 = 1300 \times 35+ 1000 \times 25](https://tex.z-dn.net/?f=P_1%20%3D%201300%20%5Ctimes%2035%2B%201000%20%5Ctimes%2025)
![P_1 =70500 Kg.m/s](https://tex.z-dn.net/?f=P_1%20%3D70500%20Kg.m%2Fs)
final momentum
![P_2 = m_aV_a + m_b V_b](https://tex.z-dn.net/?f=P_2%20%3D%20m_aV_a%20%2B%20m_b%20V_b)
![P_2 = 1300 \times 30+ 1000 \times 31.5](https://tex.z-dn.net/?f=P_2%20%3D%201300%20%5Ctimes%2030%2B%201000%20%5Ctimes%2031.5)
![P_2 =70500 Kg.m/s](https://tex.z-dn.net/?f=P_2%20%3D70500%20Kg.m%2Fs)
here initial momentum is equal to the final momentum of the car.
hence, momentum is conserved in the collision.
Alkali Metals ......................................
Answer:
They interact in a lot of ways, prey, predator, etc.