Answer:
V = 8.34m/s
Explanation:
Given that
1/2ke^2 = 1/2mv^2 ......1
Where e = 3.75cm = (3.75/100)m
e = 0.0375m
K = 500 N/m
m = 10g = 10/1000
= 0.01kg
Substitute the values into equation 1
0.5×500×(0.0375)^2 = 0.5×0.01×v^2
250×0.001395 = 0.005v^2
0.348 = 0.005v^2
v^2 = 0.348/0.005
v^2 = 69.6
V = √69.6
V = 8.34m/s
The ball launches at the speed of V = 8.34m/s
Answer:
Frequency
Explanation:
A period is the number of revolutions in a minute.

Also, frequency can be defined as number of oscillation per unit time.
Mathematically, frequency is given by the formula;

Thus, we can deduce that frequency of an oscillation is the inverse of the period (time) of oscillation.
Mathematically, period is expressed as;

This ultimately implies that, an object's period of oscillation can always be used to solve for frequency.
Answer:
The acceleration of the toboggan going up and down the hill is 8.85 m/s² and 3.74 m/s².
Explanation:
Given that,
Speed = 11.7 m/s
Coefficients of static friction = 0.48
Coefficients of kinetic friction = 0.34
Angle = 40.0°
(a). When the toboggan moves up hill, then
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula


(b). When the toboggan moves up hill, then
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula


Hence, The acceleration of the toboggan going up and down the hill is 8.85 m/s² and 3.74 m/s².
Answer:
Ethan's tangential speed is twice as Rebecca's.
Explanation:
Let
be the distance from the center of the platform to Rebecca's horse,
Rebecca's tangential speed,
the Ethan's tangential speed and
the merry-go-round angular speed. The distance from the center of the platform to Ethan's horse is
. The angular speed is related to the tangential speed by the equation:

Since the angular speed is the same for Ethan and Rebecca, we have that:

Now, solving for
we get:

It means that Ethan's tangential speed is twice as Rebecca's.
Answer:

Explanation:
Please find the image for the question as attached file.
Solution -
Given -
First of all we will calculate the velocity at point C,
As per newton's third law of motion-

Substituting the given values in above equation, we get -

Now we will determine the radius of curvature for the curve shown in the attached image

Differentiating on both the sides, we get -
meter
Acceleration on curved path

Final acceleration
