Answer:
Catapult on the ground: Normal, gravity
Catapult (I'm assuming launching marshmallow): Reaction of Force Applied
Marshmallow: Force Applied
Explanation:
This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.
Answer:
t = 1.41 sec.
Explanation:
If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.
First, we need to find the value of acceleration, which is the same for both blocks.
If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:
F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)
⇒ a = (
= g/5 m/s²
Once we got the value of a, we can use for instance this kinematic equation, and solve for t:
Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.
Answer:
The magnitude of the centripetal force to make the turn is 3,840 N.
Explanation:
Given;
radius of the cured road, r = 400 m
speed of the car, v = 32 m/s
mass of the car, m = 1500 kg
The magnitude of the centripetal force to make the turn is given as;

where;
Fc is the centripetal force

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.
What’s the problem what do u need help on
The answer would be 22.50cal