Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
1 answer:
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Answer:
14 m/s
Explanation:
We can solve the problem by using the law of conservation of energy.
At the beginning, when the ball is thrown from the ground, it has only kinetic energy, which is given by
where m = 5.9 kg is the mass of the ball and v is its initial speed.
As the ball goes up, its speed decreases, so its kinetic energy decreases and converts into gravitational potential energy. When the ball reaches its maximum height, the speed has become zero, and all the kinetic energy has been converted into gravitational potential energy, given by:
where g = 9.8 m/s^2 is the gravitational acceleration and h = 10 m is the maximum height reached by the ball.
Since we can ignore air resistance, energy must be conserved, so the initial kinetic energy must be equal to the final potential energy of the ball, so we can write:
And we can solve the equation to find v, the initial speed of the ball:
Answer:
distance of 2nd team from 1st team will be: 58.2
Direction of 2nd team from 1st team will be: 14.90 deg North of east
Explanation:
ASSUME Vector is R and makes angle A with +x-axis,
therefore component of vector R is
From above relation
Assuming base camp as the origin, location of 1st team is
away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)
location of 2nd team is at
, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)
Now position of 2nd team with respect to 1st team will be given by:
Using above values:
distance of 2nd team from 1st team will be:
Direction of 2nd team from 1st team will be:
Direction = 14.90 deg North of east