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To find the resultant vector after adding the vectors ( V1 ), ( V2 ), and ( V3 ), we can add their components separately.

Given:

• ( V1 = (-5, 0) )
• ( V2 = (0, -2) )
• ( V3 = (7, 0) )

Now, we add the x-components and y-components:

1. X-components: [ -5 + 0 + 7 = 2 ]

2. Y-components: [ 0 + (-2) + 0 = -2 ]

Now, combining the results from the x and y components, we get the resultant vector: [ (2, -2) ]

Thus, the correct choice is (2, -2).

To find the total displacement of the bee, we can break down its movements into components and then calculate the resultant vector.

1. The bee flies 2 meters east.
2. Then it flies 5 meters north.
3. Finally, it flies 3 meters east.

Now, let's sum the movements in each direction:

• Total eastward movement: (2 \text{ meters} + 3 \text{ meters} = 5 \text{ meters east})
• Total northward movement: (5 \text{ meters north})

Now we can represent the total displacement as a vector:

• Displacement = (5 \text{ meters east} + 5 \text{ meters north})

To find the magnitude of the total displacement, we can use the Pythagorean theorem:

[ \text{Displacement} = \sqrt{(5 \text{ meters})^2 + (5 \text{ meters})^2} = \sqrt{25 + 25} = \sqrt{50} \approx 7.07 \text{ meters} ]

However, since the question asks for the total displacement in terms of direction, we can say:

The total displacement is (5 \text{ meters east and } 5 \text{ meters north}).

Thus, the correct answer is:

5 meters east and 5 meters north.

To find the total displacement of the ant, we can break down its movements into components and then calculate the resultant displacement.

1. Movement Breakdown:

   • 2 meters west: This can be represented as (-2, 0).
   • 3 meters south: This can be represented as (0, -3).
   • 4 meters east: This can be represented as (4, 0).
   • 1 meter north: This can be represented as (0, 1).

2. Sum of Movements:

   • Total movement in the east-west direction: [ -2 + 4 = 2 \text{ meters east} ]
   • Total movement in the north-south direction: [ -3 + 1 = -2 \text{ meters south} ]

3. Resultant Displacement: The ant's total displacement can be represented as (2, -2), which means 2 meters east and 2 meters south.

4. Magnitude of Displacement: To find the magnitude of the displacement, we can use the Pythagorean theorem: [ \text{Magnitude} = \sqrt{(2^2) + (-2^2)} = \sqrt{4 + 4} = \sqrt{8} = 2.83 \text{ meters} ]

5. Direction: The direction can be found using the tangent function: [ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{2} = 1 \implies \theta = 45^\circ ] This means the direction is southeast.

Based on the options provided, the correct answer is:

2 meters east and 2 meters south.

To find the height of the cliff, we can use the formula for the distance fallen under the influence of gravity. The vertical motion of the object is independent of its horizontal motion.

The formula for the distance fallen due to gravity is:

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

where:

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

Given that ( t = 3 ) seconds, we can substitute the values into the formula:

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

Calculating this step-by-step:

1. Calculate ( (3 , \text{s})^2 = 9 , \text{s}^2 ).
2. Multiply by ( g ): ( 9.81 , \text{m/s}^2 \times 9 , \text{s}^2 = 88.29 , \text{m} ).
3. Now multiply by ( \frac{1}{2} ):

[ d = \frac{1}{2} \times 88.29 , \text{m} = 44.145 , \text{m} ]

Rounding this to one decimal place gives us approximately ( 44.1 , \text{m} ).

Thus, the height of the cliff is **44.1 meters

To find the horizontal distance traveled by the object, we can use the formula for horizontal distance:

[ \text{Horizontal Distance} = \text{Horizontal Speed} \times \text{Time} ]

Given:

• Horizontal Speed = 10 m/s
• Time = 3 seconds

Now, we can calculate the horizontal distance:

[ \text{Horizontal Distance} = 10 , \text{m/s} \times 3 , \text{s} = 30 , \text{meters} ]

Therefore, the horizontal distance traveled by the object is 30 meters.

So the correct answer is 30 meters.

To determine how long it takes for a golf ball to hit the ground when released from a height of 10 meters, we can use the formula for the time of free fall under the influence of gravity:

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

where:

• ( h ) is the height (10 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{2h}{g} ]

Substituting the values:

[ t^2 = \frac{2 \times 10 , \text{m}}{9.81 , \text{m/s}^2} \approx \frac{20}{9.81} \approx 2.04 ]

Now, taking the square root:

[ t \approx \sqrt{2.04} \approx 1.43 , \text{seconds} ]

Rounding this, we find that it takes approximately 1.4 seconds for the golf ball to hit the ground.

Therefore, the correct answer is 1.4 seconds.

To determine how long it takes for a golf ball to hit the ground when released from a height of 10 meters, we can use the formula for the time of free fall under the influence of gravity:

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

where:

• ( h ) is the height (10 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{2h}{g} ]

Substituting the values:

[ t^2 = \frac{2 \times 10 , \text{m}}{9.81 , \text{m/s}^2} \approx \frac{20}{9.81} \approx 2.04 ]

Now, taking the square root:

[ t \approx \sqrt{2.04} \approx 1.43 , \text{seconds} ]

Rounding this, we find that it takes approximately 1.4 seconds for the golf ball to hit the ground.

Therefore, the correct answer is **1.4 seconds

To solve this problem, we need to determine the horizontal speed required for the skydiver to cover a horizontal distance of 135 meters while falling from a height of 25 meters.

First, we calculate the time it takes for the skydiver to fall 25 meters vertically. We use the formula for the time of free fall:

[ t = \sqrt{\frac{2h}{g}} ]

where:

• ( h = 25 ) meters (height),
• ( g = 9.8 , \text{m/s}^2 ) (acceleration due to gravity).

[ t = \sqrt{\frac{2 \times 25}{9.8}} ] [ t = \sqrt{\frac{50}{9.8}} ] [ t \approx \sqrt{5.102} ] [ t \approx 2.26 , \text{seconds} ]

Next, we use the horizontal distance formula to find the horizontal speed ( v ):

[ d = v \times t ]

where:

• ( d = 135 ) meters (horizontal distance),
• ( t \approx 2.26 ) seconds (time calculated above).

Solving for ( v ):

[ v = \frac{d}{t} ] [ v = \frac{135}{2.26} ] [ v \approx 59.73 , \text{m/s} ]

Therefore, the horizontal speed with which the skydiver must leave the jump is approximately 59.73 m/s. The closest answer choice is:

59 m/s

To find the horizontal speed at which the skydiver must leave the jump, we can use the principles of projectile motion. The horizontal distance traveled (range) and the vertical drop (height) can be used to determine the required horizontal speed.

1. Calculate the time of flight: The time it takes for the skydiver to fall from a height of 25 meters can be calculated using the formula for free fall:

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

where:

• ( h ) is the height (25 m),
• ( 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 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 25 , \text{m}}{9.81 , \text{m/s}^2}} \approx \sqrt{\frac{50}{9.81}} \approx \sqrt{5.1} \approx 2.26 , \text{s} ]

2. Calculate the horizontal speed: The horizontal speed ( v ) can be calculated using the formula:

[ \text{horizontal distance} = \text{horizontal speed} \times \text{time} ]

Rearranging to solve for horizontal speed:

[ v = \frac{\text{horizontal distance}}{t} = \frac{135 , \text{m}}{2.26 , \text{s}} \approx 59.74 , \text{m/s} ]

Rounding to two significant figures, the horizontal speed is approximately ( 59 , \text{m/s} ).

Thus, the correct answer is:

**

True.

The deceleration due to gravity when an object is moving upward is equal in magnitude to the acceleration due to gravity when the object is falling downward. In both cases, the acceleration due to gravity is approximately (9.81 , \text{m/s}^2) near the Earth's surface. The direction of the acceleration is what differs: it acts downward when an object is falling and acts upward (as a deceleration) when an object is moving

True.

In the absence of air resistance, the time it takes for a projectile to rise to its maximum height is equal to the time it takes to fall back down to the original launch height. This is due to the symmetry of projectile motion under uniform gravitational acceleration.

True.

In the absence of air resistance, the speed lost by a cannonball while it is going up is equal to the speed gained while it is coming down. This is due to the conservation of energy and the symmetry of projectile motion under uniform gravitational acceleration. The cannonball will decelerate at a constant rate (due to gravity) as it ascends and will accelerate at the same rate as it descends. Therefore, the speeds at the same heights during ascent and descent will be equal.

True.

In the absence of air resistance, the speed lost by a cannonball while it is going up is equal to the speed gained while it is coming down. This is due to the conservation of energy and the symmetry of projectile motion under uniform gravitational acceleration. The cannonball will decelerate at a constant rate (due to gravity) as it ascends and will accelerate at the same rate as it descends. Therefore, the speeds at the same heights during ascent and descent will be

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.