Answer:
The radius of the circle made by the person on the merry go round is 74.55 meters
Explanation:
The given parameters are;
The force the merry go round exerts on the rider = 1000 N
The time it takes the merry go round to make one complete revolution = 15 seconds
The weight of the person = 750 N
The radius of the circle made by the person on the merry go round = r
We have;
[tex]F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r[/tex]
Where;
m = The mass of the person
v = The velocity of the person
[tex]F_c[/tex] = The centrifugal force acting on the person = 1,000 N
r = The radius of the circle made by the person on the merry go round
ω = Angular velocity = 2·π/15 rad/s
We have;
The mass of the person = The weight/(The acceleration due to gravity, g)
∴ The mass of the person = 750/9.81 ≈ 76.45 kg
By substituting the calculated and known values into the equation for the centripetal force, we have;
[tex]F_c[/tex] = m × ω² × r
1000 = 76.45 × (2·π/15)² × r
r = 1000/(76.45 × (2·π/15)²) = 74.55 m
The radius of the circle made by the person on the merry go round = r = 74.55 m.
Physical values in the real world have two components: magnitude and
Answer:
Dimension.
Explanation:
A fish swimming at a rate of .6 m/s notices a huge shark. Three seconds later, the fish is swimming at a speed of 3 m/s. What is the fish's acceleration?
0.8 m/s/s
-0.8 m/s/s
12.5 m/s/s
-12.5 m/s/s
Answer:
C
Explanation:
???
i think
Can u anser 5,6 on the picture
Answer: Number 6 is Periods
Explanation:
A pinball bangs against a bumper of a pinball machine with a speed of 0.46 m/s. If the ball has a mass of 0.058 kg, what is the ball's kinetic energy?
Answer:
either its 12 or 0.402
Explanation:
If the velocity of an object changed from 30 m/s to 60 m/s over a period of 10 seconds what would the average acceleration be ?
A 75kg bicyclist (including the bicycle), initially at rest at the top of a hill, coasts down the hill, reaching a speed of 14.6m/s at the bottom of the hill. The distance and height of the hill are shown. Neglect any friction impeding the motion and the rotational energy of the wheels. List the energy types at the initial and final time and whether work and loss (due to non-conservative forces) occur as well as the corresponding amounts of energy.
The energy type at the initial time is potential energy and the energy at the final time or position is kinetic energy.
What is the law of conservation of energy?The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.
Based on the law of conservation of mechanical energy, the formula for the change in the kinetic energy and the potential energy of the bicyclist is given as;
K.Ei + P.Ei = K.Ef + P.Ef
where;
K.Ei is the initial kinetic energy of the bicyclistK.Ef is the final kinetic energy of the bicyclistP.Ei is the initial potential energy of the bicyclistP.Ef is the final potential energy of the bicyclistThe kinetic energy of the bicyclist increases with increase in the velocity of the bicyclist while the potential energy increases with increase in the height of the bicyclist.
At the initial position when the bicyclist is at rest, the kinetic energy is zero, so the only energy at the initial position is potential energy because the height is maximum.
In addition, at the final position, the velocity of the bicyclist is maximum and the height is zero, so the only energy at the final position is kinetic energy.
Learn more about conservation of energy here: https://brainly.com/question/166559
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Shows a car travelling around a bend in the road. The car is travelling at a constant speed. There is a resultant force acting on the car. This resultant force is called the centripetal force. (i) In which direction, A, B, C or D, does the centripetal force act on the car? Tick ( ) one box. A B C D (1) (ii) State the name of the force that provides the centripetal force.
Answer:
This question is incomplete
Explanation:
This question is incomplete but the missing figure is in the attachment below.
When an object is travelling around a circular path, there is a force that tends to draw that object towards the center of the circular path and keep the object moving in the curved path, that force is called the centripetal force. From this description, it can be deduced that the direction of the centripetal force that acts on the car (in the attachment below) is D.
The name of the force that provides this centripetal force is frictional force. This is the force that prevents the car from slipping off the road; keeping it moving in the curved path.
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
- During a certain period, the angular position of a rotating object is given by: = − + , where is in radian and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the rotating object at = Sec.
The question is not complete. The complete question is :
During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.
Solution :
Given :
Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]
∴ Angular speed is [tex]$\omega = \frac{d \theta}{dt}$[/tex]
ω = 10 + 4t
And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]
α = 4
a). At time, t = 0.00 seconds :
Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]
[tex]$\theta =2(0)^2 +10(0)+5$[/tex]
= 5 rad
Angular speed is ω = 10 + 4t
ω = 10 + 4(0)
= 10 rad/s
Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]
b). At time, t = 3.00 seconds :
Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]
[tex]$\theta =2(3)^2 +10(3)+5$[/tex]
= 53 rad
Angular speed is ω = 10 + 4t
ω = 10 + 4(3)
= 22 rad/s
Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]