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Starting from rest a ball rolls down a long incline with a constant acceleration After 2 seconds it traveled 2 meters In the next second it will have travled?
In the first 2 seconds, the velocity of the ball would be given by v = at, where a is the acceleration. Given it traveled 2 meters in 2 seconds, we can use the equation s = (1/2)at^2 to find the acceleration which is 1 m/s^2. So, after 3 seconds, the ball will travel an additional 3 * 1 = 3 meters.
What is the velocity of the ball 0.6 seconds after its release?
Assuming you release it from a position of rest, you must multiply the time by the acceleration. The acceleration due to gravity near Earth's surface is approximately 9.8 meters/second squared.
What is the acceleration of the ball after 1 second?
The acceleration of the ball is 9.8 m/s^2, which is due to gravity acting on it.
A sepaktakraw is ball is heat vertically upward by a player what is its acceleration after 1 second?
The acceleration of the sepaktakraw ball after 1 second will be approximately 9.81 m/s^2, assuming the ball is under the influence of gravity. This is the acceleration due to gravity acting downwards on the ball.
How far will a rubber ball fall in ten seconds?
The distance a rubber ball falls in 10 seconds will depend on the height from which it is dropped and the acceleration due to gravity. On Earth, neglecting air resistance, the general equation to calculate the distance fallen is: distance = 0.5 * acceleration due to gravity * time^2.
Starting from rest a ball rolls down a long incline with a constant acceleration After 2 seconds it traveled 2 meters In the next second it will have travled?
In the first 2 seconds, the velocity of the ball would be given by v = at, where a is the acceleration. Given it traveled 2 meters in 2 seconds, we can use the equation s = (1/2)at^2 to find the acceleration which is 1 m/s^2. So, after 3 seconds, the ball will travel an additional 3 * 1 = 3 meters.
After 2 seconds its velocity is 9 meters per second what is the accleration of a ball?
If the ball started off stationary and the acceleration is the same for the 2 seconds, then the acceleration of the ball is:Speed / time = acceleration9 (m/s) / 2 (s) = 4.5 (m/s/s)Hope this helps.
What is the velocity of the ball 0.6 seconds after its release?
Assuming you release it from a position of rest, you must multiply the time by the acceleration. The acceleration due to gravity near Earth's surface is approximately 9.8 meters/second squared.
What is the acceleration of the ball after 1 second?
The acceleration of the ball is 9.8 m/s^2, which is due to gravity acting on it.
A sepaktakraw is ball is heat vertically upward by a player what is its acceleration after 1 second?
The acceleration of the sepaktakraw ball after 1 second will be approximately 9.81 m/s^2, assuming the ball is under the influence of gravity. This is the acceleration due to gravity acting downwards on the ball.
How far will a rubber ball fall in ten seconds?
The distance a rubber ball falls in 10 seconds will depend on the height from which it is dropped and the acceleration due to gravity. On Earth, neglecting air resistance, the general equation to calculate the distance fallen is: distance = 0.5 * acceleration due to gravity * time^2.
Can you give two examples of acceleration?
A car increasing its speed from 0 to 60 mph in 10 seconds is an example of acceleration. A ball thrown into the air, moving faster and faster as gravity pulls it downward, is also experiencing acceleration.
What is the height of a ball dropped from 80 meters after 3 seconds?
The height of the ball after 3 seconds can be calculated using the formula for free fall: ( h = h_0 - \frac{1}{2} g t^2 ), where ( h_0 ) is the initial height (80 meters), ( g ) is the acceleration due to gravity (approximately 9.81 m/s²), and ( t ) is the time in seconds. After 3 seconds, the height is ( h = 80 - \frac{1}{2} \times 9.81 \times (3^2) ), which simplifies to ( h = 80 - 44.145 ). Therefore, the height of the ball after 3 seconds is approximately 35.855 meters.
Can a body have constant acceleration and zero velocity?
Since the derivative of velocity is acceleration, the answer would be technically 'no'. Here is why: v = 0 v' = 0 = a Or in variable form... v(x) = x v(0) = 0 v'(0) = 0 = a You can "trick" the derivative into saying that v'(x) = 1 = a (since the derivative of x = 1) and then stating v'(0) = 1 = a... but that is not entirely correct. Acceleration is a change over time and is measured at more then one point (i.e. the acceleration of this body of matter is y from time 1 to 5) unless using derivatives to form the equation of the acceleration line/curve. If an object has a constant acceleration of 1, then the velocity is constantly increasing over that time. Using the equation discussed above and looking at acceleration over time, at 0 seconds, acceleration is 0 and so is velocity, but from 0-1 seconds acceleration is 1 and velocity is 1 as well. 0-2 seconds, acceleration is 1, but velocity would be 2 (at the end of 2 seconds).
What force would be needed to produce an acceleration of 1 ms-2 on a ball of mass 1 kg?
The force needed would be equal to the mass of the ball multiplied by the acceleration, so 1 kg * 1 m/s^2 = 1 Newton.
How do you calculate acceleration between 6 and 9 seconds?
To calculate acceleration between 6 and 9 seconds, you need to find the change in velocity during that time interval and then divide it by the time taken. The formula for acceleration is acceleration = (final velocity - initial velocity) / time. Plug in the velocities at 6 seconds and 9 seconds into the formula to get the acceleration.
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant before it hits the ground?
To calculate the velocity of the ball just before it hits the ground, we can use the equation of motion: velocity = acceleration x time. The acceleration due to gravity is approximately 9.8 m/s^2. Given the time of 3.0 seconds, we can plug these values into the equation to find the velocity. Therefore, the velocity of the ball just before it hits the ground is 29.4 m/s.
