saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of earth is 6.4×10^6 meter

Answer:.

Required velocity = 6.26ms^-1

Explanation:

Given,

Distance, s = 450m

Time, t = 2 sec

Step 1. We obtain the distance covered within the given time under gravitational acceleration, g = 9.8ms^-2

S = ut + (1/2)gt^2. :; u = 0

: S = (1/2)gt^2

=(1/2) (9.8)(2^2)

= 19.6m

Step 2 :

We obtain the velocity using the formula.

V^2 = u^2 + 2gs.

Where u is initial velocity, v is final/ required velocity

Again u = 0

: V^2 = 2 (9.8)(19.6)

= 39.2

: V = 6.26ms^-1

Answer:

the awnser is A

Explanation:

Brainliest Please

Answer:

the second derivative of y with respect to time gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string

Explanation:

This is because the transverse wave movement of particles take place in direction 90° to direction of movement of the wave (x) itself, so second derivative of y with respect to time (t)is what will be required

Answer:

0.123 m.

Explanation:

From Hook's law,

The potential energy of the book = the energy stored in the spring.

mgh = 1/2ke².................. Equation 1

Where m= mass of the book, g = acceleration due to gravity, h = height, k = spring constant of the spring, e = distance of compression.

make e the subject of the equation

e = √(2mgh/k).................. Equation 2

Given: m = 1.3 kg, h = 0.8 m, k = 1350 N/m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×1.3×0.8×9.8/1350)

e = √(20.384/1350)

e = √(0.0151)

e = 0.123 m.

Answer:

0.015m (downwards)

Explanation:

When the book is dropped on the top of the spring at that height, the potential energy () of the book is converted to elastic energy () on the spring thereby causing a compression on the spring. i.e

=

But;

The potential energy of the mass (book), is the product of the mass(m) of the book, the height(h) from which it was dropped and the acceleration due to gravity (g). i.e

= - m x g x h         [the -ve sign shows a decrease in height as the mass (book) drops]

Also;

The elastic energy () of compression of the spring is given by

= x k x c

Where;

c = compression length of the spring

k = the spring's constant

Substitute these values of and E into equation (i) as follows;

- m x g x h = x k x c             ----------------(ii)

From the question;

m =  1.3kg

h =  0.8m

Take g = 10m/s²

k =  1350N/m

Substitute these values into equation (ii) as follows;

- 1.3 x 10 x 0.8 = x 1350 x c

- 10.4 = 675c

Solve for c;

c = - 0.015 m          [The negative sign shows that the spring actually compresses]

Therefore, the maximum distance the spring will be compressed is 0.015m (downwards of course).

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

Put the value into the formula

Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Final answer:

The question is about determining the horizontal displacement of a projectile based on the given initial speed and projection angle and the height of the launch. This can be calculated using the equations of motion, specifically those pertaining to projectile motion.

Explanation:

In this problem, we're dealing with projectile motion. The stone being thrown is the projectile in this case. The horizontal displacement, also known as range, of a projectile can be defined using the formula: range = (initial speed * time of flight) * cosθ, where θ is the angle of projection. The initial speed is given as 11.5 m/s and the angle as 50 degrees. Now, we need to calculate the time of flight. This can be found by the formula: time of flight = (2 * initial speed * sinθ) / g. Considering g, the acceleration due to gravity, as 9.8 m/s², we can find the time of flight and thus calculate the range. Always remember that while the vertical motion of a projectile is affected by gravity, the horizontal motion remains constant.

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The benefits of learned optimism that have been found in research are Fewer health problems, Making more money, and a lower divorce rate. The correct option is B.

Learned optimismhas been associated with numerous benefits in research, including fewer health problems, making more money, and a lower divorce rate. Optimistic people tend to have better physical and mental health, which leads to fewer health problems. Additionally, optimistic people tend to be more successful in their careers and finances, which can lead to higher income and better financial stability. Finally, optimistic people tend to have better relationships, including lower divorce rates, as they are better able to handle conflicts and maintain positive attitudes toward their partners.

In summary, learned optimism has a range of benefitsfor individuals, including better physical and mental health, greater success in work and education, better relationships with others, and improved resilience. These benefits make learned optimism an important skill for individuals to develop in order to lead happier, healthier, and more successful lives.

To learn about  pessimism  click:

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