Answer:
The correct answer is option 'b':
Explanation:
As we know that the image formed by a plane mirror is at same distance behind the mirror as the object is placed in front of the mirror.
Hence we can conclude that the speed of the image as it moves towards the mirror is equal to the speed of the object as it moves towards the mirror.
Thus the relative speed of the object and the image can be calculated as
where
is the speed of the image towards mirror
is the speed of the object towards mirror
Applying values we obtain the relative speed as
Answer:
Explanation:
Given:
angular speed of rotation of friction-less platform,
moment of inertia with extended weight,
moment of inertia with contracted weight,
<u>Now we use the law of conservation of angular momentum:</u>
The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.
Answer:
a) f = 453.3Hz
b) f = 443.1Hz
c) f = 420Hz
Explanation:
First of all we need to know the positions and velocities of both vehicles.
For the car:
Xc(-5) = 0m; Vc(-5) = 0m/s
; Vc(5)=2.5*5 = 12.5m/s
; Vc(10)=2.5*10=25m/s
For the truck:
Xt(-5)=10*(-5) = -50m; Vt(5)=10m/s
Xt(5)=10*5 = 50m; Vt(5)=10m/s
Xt(10)=10*10=100m; Vt(10)=10m/s
Now for part a) t=-5s. The truck is behind the car, so:
Now for part b) t=5s. The car is behind the truck, so:
Now for part b) t=10s. The truck is behind the car, so:
The car can travel 220.27 ft in 4.61 s of time.
Given that a sports car accelerates from a standing start to 65 mi/hr.
So the initial velocity = 0 ft/s
The final velocity = 65 mi/hr.
We convert the final velocity from mi/hr. to ft/s.
1 mi/hr = 5280/3600 ft/s
= 1.47 ft/s
Therefore final velocity = 65 x 1.47 = 95.55 ft/s
The time taken by the car = 4.61 s
Now the average acceleration of the car = final velocity - initial velocity / time taken = 95.55 - 0/ 4.61 s = 20.73
The displacement of the car , ,
where u is the initial velocity of the car, t, the time taken by the car and a, the average acceleration.
So, displacement = 0 x 4.61 + 1/2 x 20.73 x
= 220.27 ft
Hence the distance travelled by the sports car in 4.61s is 220.27 ft.
Learn more about the displacement at brainly.com/question/17006213
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