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ohaa [14]
3 years ago
15

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground. Calculate the velocity of the acorn

just before it reaches the ground (rounding your answer to the integer) and its kinetic energy when hitting the ground (rounding your answer to the nearest tenth)
Physics
1 answer:
sdas [7]3 years ago
5 0

Answer:

The velocity of the acorn just before it reaches the ground is 19 m/s

The kinetic energy when hitting the ground is 3.1 J

Explanation:

Given;

mass of the acorn, m = 0.017 kg

height of fall, h = 18.5 m

Apply the law of conservation of mechanical energy;

mgh = ¹/₂mv²

gh = ¹/₂v²

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 18.5)

v = 19 m/s

Thus, the velocity of the acorn just before it reaches the ground is 19 m/s

Now, determine the kinetic energy when hitting the ground;

K.E = ¹/₂mv²

K.E = ¹/₂(0.017)(19)²

K.E = 3.09 J

K.E = 3.1 J

Therefore, the kinetic energy when hitting the ground is 3.1 J

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Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t
ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body       Mass (gram)     Engine      Force (newtons)

1                   500                 1                     25

2                  1500                2                    20

3                  750                  3                    30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

 20    = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

                      $=\frac{200}{5}$

                      $= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $ 40 \ m/s^2$ at the start of the launch.

8 0
3 years ago
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

kx_1 = mg

The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

kx_2 = m\frac{g}{6}

kx_2 = \frac{1}{6} mg

kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

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Gala2k [10]

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