High speed stroboscopic photographs show that the head of a 244 g golf club is traveling at 57.6 m/s just before it strikes a 45
1 answer:
You might be interested in
Answer:
None of the above forces on air drag on him is equal to his weight
Explanation:
In the velocity-time graph,the gradient of the curve where it is flatten shows the parachutist reaches the terminal velocity when it reaches terminal velocity which means the parachutist reaches constant velocity or speed,indicating that the acceleration of free fall(g) is zero.And according to the resultant force formula weight - air drag= mass*acceleration. so when accelerate is zero,resultant force is zero. And hence the equation will be like this: weight= air drag
The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.
dr/dt = 4 ft/sec
after 12 seconds, the radius becomes =
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt
dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec