Which is the best example of Newton's Second Law of Motion?
A student has on roller skates and he decides to push against the railing. He immediately begins to roll backwards.

A small, lightweight ball and a large, heavy ball are dropped off of a roof. They both strike the ground at the same time.

A baseball player hits a baseball that is pitched to him. The ball immediately soars back in the direction of the pitcher.

You are riding in a car that makes a quick right turn. You immediately slide across the back seat.

Answers

Answer 1
The baseball one because it shows that the ball "soars" (accelerates) due to the force applied F=ma
Answer 2
Final answer:

The baseball player hitting a baseball is the best example of Newton's Second Law of Motion, as the force exerted on the ball causes it to move in the opposite direction with the same magnitude.

Explanation:

The best example of Newton's Second Law of Motion is the baseball player hitting a baseball that is pitched to him. When the player hits the ball, the force exerted on the ball causes it to move in the opposite direction with the same magnitude. This is known as the law of action and reaction, which is a part of Newton's Second Law of Motion.

Another example of Newton's Second Law is a student on roller skates pushing against a railing and immediately rolling backwards. The force exerted by the student on the railing causes an equal but opposite force on the student, resulting in the backward motion.

Lastly, the example of a car making a quick right turn and the passenger sliding across the back seat can be considered an example of Newton's Second Law. The passenger tends to continue moving in a straight line due to inertia but is pushed to the side due to the force of the car turning.

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Related Questions

A car initially moving at 9 m/s begins accelerating at its maximum acceleration of 4 m/s2. if the car maintains this acceleration, how much time will it take to cover 200 m?

Answers

The answer is in attachment.

A constant force of f = 47i + 30j moves an object along a vector d = 13i - 8j, where units are in pounds and feet. find the work done.

Answers

Answer:

The work is  371 lb ft

Explanation:

The work can be calculated with yhe scalar product between the force and the displacement:

[tex] W = F.d [/tex]

The scalar product is obtained multiplying component to comoponent and summing the terms:

[tex] W [/tex] = <47, 30> · <13, -8>  [lb ft]

[tex] W [/tex] = 47*13  + 30*(-8) [lb ft]

Finally:

W = 371 [lb ft]

A motorist drives along a straight road at a constant speed of 15.0 m/s. Just as she passes a parked motorcycle police offi cer, the offi cer starts to accelerate at 2.00 m/s2 to overtake her. Assuming that the offi cer maintains this acceleration, (a) determine the time interval required for the police offi cer to reach the motorist. Find (b) the speed and (c) the total displacement of the offi cer as he overtakes the motorist.

Answers

The answer is in attachment.

Answer:

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²    

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

What happens to the brightness of bulb a when the switch is closed and bulb b lights up?

Answers

The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.

When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.

Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.

So the light of "bulb a" will be dimmer.

There is a pattern of current flow. Based on the above scenario,  Option b is the correct answer as bulb B will be brighter than before (Image attached).

What does this implies?

Based on the current flowing through bulb A  is said to remains the same in both scenario. It is the current through bulb B that changes when the switch  are said to be closed.

There is something that happens to the brightness of a bulb when the switch is said to be closed. Note that when the switch is closed, the light bulb often operates because of the current flows via the circuit. The bulb (B) tends to glows at its full brightness because it receives more volts.

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A boat can travel 48 miles upstream in 4 hours. the return trip takes 3 hours. find the speed of the boat in still water and the speed of the current.

Answers

Upstream velocity=48/4=12mph
Downstrean velocity=48/3=16mph

On a planimetric map, the _____ is used to determine the actual distance between two points.

Answers

The answer in this question is scale. The scale is used to determine the actual distance between two points in a planimetric map. This map only represents the horizontal position of features. This is also called by others as the line map.  This is usually consisted of man-made and natural features. These are from aerial shots by the use of modern technology. This map’s common features are street and water centerlines, sidewalks, culverts, utility lines, building footprints and vegetation. Sometimes, there are also other useful data attached in this map like titles of the properties, owner, assessed value, etc. It is because of these, this map becomes a storage of valuable data which can be easily and quickly accessed. 

Final answer:

The scale on a planimetric map is used to determine the actual distance between two points by converting measured map distances to real-world distances.

Explanation:

On a planimetric map, the scale is used to determine the actual distance between two points. By measuring the distance between two points on the map with a ruler and then using the map scale to find the real distance on the land surface, it is possible to convert map distances to real-world distances. For example, if you measure 10 centimeters between two points on a map with a scale of 1:10,000, you multiply the measured distance by the scale factor, resulting in an actual distance of 100,000 centimeters on the Earth's surface.

It is important to note that large-scale maps, like a standard topographic map, typically have less distance distortion, making it easier to measure straight line distances. However, when measuring distances along curved features, techniques such as using multiple straight-line segments or tools like an opisometer may be employed for greater accuracy. Map scales can be complex, as they can vary with latitude due to the Earth's curvature, particularly on maps that cover large areas such as equidistant projections which are true-to-scale only along meridians.

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