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3 years ago

. A ball is thrown downward at a speed of 20 m/s. Choosing

Physics

1 answer:

a_sh-v [17]3 years ago

3 0

The final velocity is $v = 20 - gt$

The distance traveled by the ball at time t is $y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is $0 = (20)^2 - 2g(y-y_0)$

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

$v = 20 - gt$

The distance traveled by the ball at time t;

$y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is calculated as;

$v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)$

Learn more here: brainly.com/question/16878713

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The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25

Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0

3 years ago

The action that has the ability to change the motion of an object is

Troyanec [42]

The answer to this question is force

3 0

3 years ago

A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must

nika2105 [10]

Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.

The moment due to cat = 5.3*2 = 10.6 kg.m
The moment due to bowl = 2.5*2 = 5 kg.m
The unbalanced moment = 10.6 - 5 = 5.6 kg.m

Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)

7 0

3 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given