Answer:
The answer is
<h2>10 m/s²</h2>
Explanation:
To find the acceleration of an object given the force and mass we use the formula
<h3>

</h3>
From the question
mass of object = 50 kg
force = 500 N
So the acceleration is
<h3>

</h3>
We have the final answer as
<h3>10 m/s²</h3>
Hope this helps you
We have that the instantaneous velocity of the
shuttlecock when it hits the ground is

From the question we are told
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration is mathematically given as

Where
acceleration is still -9.81 m/s2,
Hence,

Therefore

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let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is 
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.
The X and Y components are as follows;
1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s
2. X = 12 * -cos 34 = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s
3. X = 8 * -cos 90 = 0 m/s; Y = 12 -sin 90 = -8 m/s
4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) = -19.3 m/s
<h3>What are the horizontal and vertical components of the vectors?</h3>
The horizontal and vertical components of the velocities are given as follows:
- Horizontal component, X = x cos θ
- Vertical component, Y = y sin θ
1. 35 m/s at 57° from x-axis
X = 35 * cos 57 = 19. 1m/s
Y = 35 * sin 57 = 29.4 m/s
2. 12m/s at 34° S of W
X = 12 * -cos 34 = -10 m/s
Y = 12 * -sin 34 = -6.7 m/s
3. 8 m/s at South
X = 8 * -cos 90 = 0 m/s
Y = 12 -sin 90 = -8 m/s
4. 20 m/s at 275° from x-axis
X = 20 * cos 75 = 5. 2m/s
Y = 20 * (-sin 75) = -19.3 m/s
In conclusion, the X and Y components are found by taking cosines and sine of the angles.
Learn more about horizontal and vertical components at: brainly.com/question/26446720
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