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2 years ago

Hector drew the diagram below to illustrate projectile motion.

Physics

2 answers:

Serjik [45]2 years ago

6 0

<u>Answer </u>

He should make the arrow for “Path” curve downward.

<u>Explanation </u>

The force of gravity is usually directly downward. So from the diagram, it is correctly labelled.

If the object was given a horizontal force, the direction of the inertial is also correct. Inertial is the force the resist the change of state of motion.

What Hector should change is the path followed by the object. It will be a curve not a straight line as it is drawn.

The correct answer is He should make the arrow for “Path” curve downward.

Eduardwww [97]2 years ago

5 0

Let us review the answers given.
1. He should switch the labels "Gravity" and "Inertia".
These labels are correct, so no change is required.

2. He should make the arrow for "Path" curve downward.
This statement is correct.

3. He should switch the labels "Inertia" and "Path".
No change is required.

4. He should make the arrow for "Inertia" curve downward.
This statement is incorrect.

Answer:
Hector should make the arrow for "Path" curve downward as shown in the figure.

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2 years ago

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Answer:

x = 0.54 m

y = 0.058 m

Explanation:

m = mass of the bullet = 16 g = 0.016 kg

v = speed of bullet before collision = 240 m/s

M = mass of the pendulum = 3.6 kg

L = length of the string = 2.5 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.016) (240) = (0.016 + 3.6) V

V = 1.062 m/s

Using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

(9.8) h = (0.5) (1.062)²

h = 0.058 m

y = vertical displacement = h = 0.058 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x = sqrt(2.5² - (2.5 - 0.058)²)

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2 years ago

How tall in cm is 5.4 feet

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Explanation:

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3 years ago

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Answer:

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$s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m$

∴ The engineer applied the brakes 500 m from the station

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3 years ago