Answer:
a. 1.64 m/s²
Explanation:
Centripetal acceleration is the square of tangential velocity divided by the radius.
a = v²/r
First, convert km/h to m/s.
30.0 km/h (1000 m/km) (1 h / 3600 s) = 8.33 m/s
Find the acceleration.
a = (8.33 m/s)² / (42.4 m)
a = 1.64 m/s²
The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.
Therefore,
In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,
Substitute the value of t from equation (1).
Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.
The shot put was thrown with a speed 15.02 m/s.
Answer:
one at the edge
Explanation:
The relation between the linear velocity and the angular velocity is given by
v = r x ω
Where, v be the linear velocity, ω be the angular velocity and r be the radius of the circular path.
As the angular velocity is constant, thus, the linear velocity depends on the radius of circular path.
So, the horse which is near to the edge has maximum radius of circular path in which it is rotating. So, the horse which is at the edge of the merry go round has maximum linear speed.
Answer:
1)Observe a phenomenon
2)Ask a question/ start inferring
3)Form a hypothesis
4)Create an experiment
5)Collect data
6)Compare results
7)Analyze
8)Report findings
9)Compare with other experiments
I would said A is the best option if i’m wrong sorry
