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White raven [17]
2 years ago
15

Find the speed of a rock which is thrown off the top of a 20 m tall building at 15 m/s when it makes contact with a bird which i

s flying at an altitude of 10 m above the ground.
Physics
2 answers:
Feliz [49]2 years ago
6 0

Answer:

v_f=20.52\frac{m}{s}

Explanation:

In this case the rock is under an uniformly accelerated motion. Thus, we use the kinematic equation that relates the final speed of an object with its initial speed, its acceleration and its traveled distance.

v_f^2=v_0^2+2a\Delta y\\v_f^2=v_0^2+2a(y_2-y_1)\\v_f^2=(15\frac{m}{s})^2+2(9.8\frac{m}{s^2})(20m-10m)\\v_f=\sqrt{421\frac{m^2}{s^2}}\\v_f=20.52\frac{m}{s}

Dominik [7]2 years ago
6 0
<h2>Answer:</h2>

20.62m/s

<h2>Explanation:</h2>

Using one of the equations of motion given by;

v² = u² + 2as     -------------------------(i)

where;

v = final velocity of the object (rock)

u = initial velocity of the object (rock) = 15m/s

a = acceleration due to gravity of the rock = g = +10m/s² (since the direction of the rock is downwards in the direction of gravity towards the flying bird).

s = distance traveled by the rock.

Notice that the rock is thrown from a 20m height to hit a bird flying at an altitude of 10m. This means that the distance (s) travelled by the rock is;

20m - 10m = 10m

Substituting the values of u, a and s into equation (i) gives;

v² = 15² + (2 x 10 x 10)

v² = 225 + 200

v² = 425

v = √425

v = 20.62m/s

Therefore the speed of the rock when it makes contact with the bird is 20.62m/s

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A ball with mass of 0.050 kg is dropped from a height of h1 = 1 .5 m. It collides with the floor, then bounces up to a height of
Iteru [2.4K]

Answer:

Explanation:

Impulse of reaction force of floor = change in momentum

Velocity of impact = √ 2gh₁

= √ 2 x 9.8 x 1.5 = 5.4 m /s.

velocity of rebound = √2gh₂

= √ 2x 9.8 x 1

= 4.427 m / s.

Initial momentum = .050 x 5.4 = .27 kg m/s

Final momentum = .05 x 4.427 = .22 kg.m/s

change in momentum = .27 - .22 = .05 kg m/s

Impulse = .05 kg m /s

Impulse = force x time

force = impulse / time

.05 / .015 = 3.33 N.

kinetic energy = 1/2 m v²

Initial kinetic energy = 1/2 x .05 x 5.4²

= 0.729 J

Final Kinetic Energy =1/2 x .05 x 4.427²

= 0.489 J

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4 0
3 years ago
A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa
gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

F-mg=ma

Solving for the force F, we obtain that:

F=ma+mg=m(a+g)

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0
3 years ago
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