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True.

The statement describes acceleration due to gravity as 10 m/s², which means that an object in free fall will increase its velocity by 10 meters per second every second. Therefore, if an object is accelerating at this rate, it is indeed traveling at 10 meters per second every second (m/s/s).

False.

In the context of a ball being thrown upwards, the acceleration due to gravity is typically -9.81 m/s² (downward). If you are referring to a scenario where the ball is accelerating upwards at 5 m/s², that would imply an additional force is acting on it, such as a propulsion force. However, in free fall or when only considering the effects of gravity, the ball would be accelerating downwards at approximately 9.81 m/s².

If the statement is meant to imply that the ball is accelerating upwards at 5 m/s², it would need to specify that there is an external force acting on it. Otherwise, the statement is misleading.

False.

When a ball is thrown straight up, it experiences an acceleration due to gravity, which is approximately -9.81 m/s² (downward) regardless of its motion. At the moment it leaves your hand, while its initial velocity may be positive (upward), its acceleration is not 0; it is still subject to the force of gravity acting

The statement that is not true of the motion of projectiles is:

A projectile with an upward component of motion will have an upward component of acceleration.

This statement is incorrect because, regardless of the direction of the projectile's motion, the acceleration due to gravity always acts downward at a constant value of approximately 9.8 m/s². Therefore, even when a projectile is moving upward, its acceleration is downward, not upward.

The statement that is not true of the motion of projectiles is:

A projectile with an upward component of motion will have an upward component of acceleration.

This statement is incorrect because, regardless of the direction of the projectile's motion, the acceleration due to gravity always acts downward at a constant value of approximately 9.8 m/s². Therefore, even when a projectile is moving upward, its acceleration is downward,

A scalar quantity is one that is described by a magnitude alone and does not have a direction. Among the options provided, the only statement that represents a scalar description is:

60 minutes

The other options (20 meters East, 3.5 m/s/s south, and 9.8 m/s North) all include directional components, making them vector quantities.

In the context of physics, vectors are quantities that have both magnitude and direction. Let's analyze the options provided:

1. Displacement: This is a vector quantity because it has both magnitude (the distance between two points) and direction (the direction from the starting point to the endpoint).

2. Acceleration: This is also a vector quantity because it describes the rate of change of velocity, which includes both magnitude (how much the velocity changes) and direction (the direction of the change in velocity).

3. Distance traveled: This is a scalar quantity because it only has magnitude (the total length of the path traveled) and does not have a direction.

Based on this analysis, the quantities that are vectors are:

• Displacement
• Acceleration

So, the correct answer is: Displacement and acceleration.

To determine how far the baseball drops during its flight, we can use the following steps:

1. Calculate the time of flight: Since the baseball is thrown horizontally, we can find the time it takes to travel the horizontal distance using the formula:

   [ \text{time} = \frac{\text{distance}}{\text{velocity}} ]

   Given:

   • Horizontal distance = 18 m
   • Horizontal velocity = 44 m/s

   [ \text{time} = \frac{18 , \text{m}}{44 , \text{m/s}} \approx 0.409 , \text{s} ]

2. Calculate the vertical drop: The vertical drop can be calculated using the formula for the distance fallen under constant acceleration due to gravity:

   [ d = \frac{1}{2} g t^2 ]

   where:

   • ( g ) (acceleration due to gravity) ≈ 9.81 m/s²
   • ( t ) is the time of flight we just calculated.

   Plugging in the values:

   [ d = \frac{1}{2} \times 9.81 , \text{m/s}^2 \times (0.409 , \text{s})^2 ]

   [ d \approx \frac{1}{2} \times 9.81 \times 0.167 \approx 0.82 , \text{m} ]

Thus, the ball drops approximately 0.82 m during its flight.

The correct answer is 0.82 m.

To find out how long the baseball stays in the air, we can use the formula for horizontal motion:

[ \text{Distance} = \text{Velocity} \times \text{Time} ]

We can rearrange this formula to solve for time:

[ \text{Time} = \frac{\text{Distance}}{\text{Velocity}} ]

Given:

• Distance = 18 m
• Velocity = 44 m/s

Now, substituting the values into the formula:

[ \text{Time} = \frac{18 , \text{m}}{44 , \text{m/s}} \approx 0.409 , \text{s} ]

Rounding this to two decimal places gives us approximately 0.41 seconds.

Thus, the ball stays in the air for 0.41 sec.

When a pebble is dropped, it is subject to the acceleration due to gravity, which is approximately (9.8 , \text{m/s}^2) downward. To find the velocity of the pebble after 3 seconds, you can use the formula for velocity under constant acceleration:

[ v = u + at ]

where:

• ( v ) is the final velocity,
• ( u ) is the initial velocity (0 m/s, since the pebble is dropped),
• ( a ) is the acceleration due to gravity ((9.8 , \text{m/s}^2)),
• ( t ) is the time (3 seconds).

Plugging in the values:

[ v = 0 + (9.8 , \text{m/s}^2)(3 , \text{s}) = 29.4 , \text{m/s} ]

So, the pebble will be traveling at (29.4 , \text{m/s}) downward after 3 seconds. The correct answer is:

29.4 m/s

To determine how long it took for the compass to strike the bottom after being dropped from a height of 240 meters, we can use the formula for the time of free fall under the influence of gravity:

[ d = \frac{1}{2} g t^2 ]

where:

• (d) is the distance fallen (240 meters),
• (g) is the acceleration due to gravity (approximately (9.81 , \text{m/s}^2)),
• (t) is the time in seconds.

Rearranging the formula to solve for (t):

[ t^2 = \frac{2d}{g} ]

Substituting the values:

[ t^2 = \frac{2 \times 240 , \text{m}}{9.81 , \text{m/s}^2} ]

Calculating the right side:

[ t^2 = \frac{480}{9.81} \approx 48.94 ]

Now, taking the square root to find (t):

[ t \approx \sqrt{48.94} \approx 7.0 , \text{seconds} ]

Thus, the time it took for the compass to strike the bottom is approximately 7.0 seconds.

The correct answer is 7.0 seconds.

To find the height of the bridge from which the marble was dropped, we can use the formula for the distance fallen under the influence of gravity:

[ d = \frac{1}{2} g t^2 ]

where:

• ( d ) is the distance (height of the bridge),
• ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )),
• ( t ) is the time in seconds (3 seconds in this case).

Plugging in the values:

[ d = \frac{1}{2} \times 9.81 , \text{m/s}^2 \times (3 , \text{s})^2 ] [ d = \frac{1}{2} \times 9.81 \times 9 ] [ d = \frac{1}{2} \times 88.29 ] [ d = 44.145 , \text{m} ]

Rounding to one decimal place, the height of the bridge is approximately ( 44.1 , \text{m} ).

So, the correct answer is 44.1 m.

To determine the speed of a free-falling object under the influence of gravity, we can use the formula for the velocity of an object in free fall:

[ v = v_0 + g \cdot t ]

where:

• ( v ) is the final velocity,
• ( v_0 ) is the initial velocity (0 m/s in this case),
• ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )),
• ( t ) is the time in seconds.

1. After 1 second: [ v = 0 + 9.8 \cdot 1 = 9.8 , \text{m/s} ]

2. After 2 seconds: [ v = 0 + 9.8 \cdot 2 = 19.6 , \text{m/s} ]

So, the object will be traveling at 9.8 m/s after 1 second and 19.6 m/s after 2 seconds.

The correct answer is: **9.8 m/sec after 1 second and 19.6 m/sec after 2 seconds

It seems like you're asking for a more detailed or extended explanation regarding the application of a certain concept or tool. However, your request is a bit unclear. Could you please provide more context or specify what application you are referring to? This will help me give you a more accurate and comprehensive response.

To determine the speed of a freely falling object after 10 seconds, we can use the formula for the speed of an object in free fall:

[ v = g \cdot t ]

where:

• ( v ) is the final speed,
• ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )),
• ( t ) is the time in seconds.

Plugging in the values:

[ v = 9.81 , \text{m/s}^2 \cdot 10 , \text{s} = 98.1 , \text{m/s} ]

Rounding this, we find that the speed is approximately ( 100 , \text{m/s} ).

Therefore, the correct answer is:

100 m/s

To solve this problem, we can use the equations of motion under constant acceleration due to gravity.

1. Finding the time of fall (t): We can use the equation for distance fallen under constant acceleration: [ d = \frac{1}{2} g t^2 ] where:

• (d = 20 , \text{m}) (the distance fallen),
• (g = 10 , \text{m/s}^2) (acceleration due to gravity).

Rearranging the equation to solve for (t): [ 20 = \frac{1}{2} \cdot 10 \cdot t^2 ] [ 20 = 5t^2 ] [ t^2 = \frac{20}{5} = 4 ] [ t = \sqrt{4} = 2 , \text{s} ]

2. Finding the final velocity (v): We can use the equation for final velocity under constant acceleration: [ v = g t ] Substituting the values we have: [ v = 10 \cdot 2 = 20 , \text{m/s} ]

Since the ball is falling downwards, we can consider the velocity to be negative in the context of direction: [ v = -20 , \text{m/s} ]

Thus, the final results are:

• The velocity at which it hits the ground is (v = -20 , \text{m/s}).
• The time it took for the ball to fall is (t = 2 , \text{s}).

The correct answer from the choices provided is: v = -20 m/s, t = 2 s.

As an object falls freely on Earth, its velocity increases due to the acceleration caused by gravity. The acceleration due to gravity is approximately (9.81 , \text{m/s}^2) and remains constant near the Earth's surface, so the acceleration does not increase as the object falls.

Therefore, the correct answer is:

**velocity increases

The date on which an act comes into effect, as well as any information regarding its commencement, may be

During the entire motion of the ball tossed vertically upward, the acceleration is always directed downward. This is due to the force of gravity acting on the ball, which pulls it downward regardless of whether it is moving upward or downward.

So, the correct answer is:

directed downward.

1. If you drop a feather and a coin at the same time in a tube filled with air, the coin will reach the bottom of the tube first. This is because the feather is affected more by air resistance than the coin, which is denser and less affected by air resistance.

2. For the ball tossed vertically upward, the acceleration of the ball is always directed downward. This is due to the force of gravity acting on the ball, which pulls it back toward the Earth regardless of its motion (upward or downward).

In a tube filled with air, the coin will reach the bottom first. This is because the feather is much lighter and has a larger surface area relative to its weight, which causes it to be affected more by air resistance. In contrast, the coin is denser and less affected by air resistance, allowing it to fall faster.

If the tube were in a vacuum (with no air), both the feather and the coin would fall at the same rate and reach the bottom at the same time due to the effects of gravity being the same on both objects. However, in the presence of air, the coin will reach the bottom first.

When no air resistance acts on a projectile, its horizontal acceleration is zero.

In the absence of air resistance, the only force acting on the projectile is gravity, which acts vertically downward. This means that there is no horizontal force acting on the projectile, resulting in zero horizontal acceleration. The horizontal component of the projectile's motion remains constant.

In the absence of air resistance, the speed of the ball when it is caught will be the same as the speed it had when thrown upwards. This is due to the principle of conservation of energy and the symmetry of projectile motion. When the ball is thrown upwards, it slows down due to gravity until it reaches its highest point, and then it accelerates back down to the original height, regaining the same speed it had when it was thrown.

Therefore, the correct answer is:

**the same as the speed it had when thrown upwards

The extended title appears below the abbreviated title and outlines the act's scope or objectives.