A boy slides a book across the floor, using a force of 5 N over a distance of 2
m. What is the kinetic energy of the book after he slides it? Assume there is
no friction.
A. 5 J
B. 10 J
C. 20 J
D. 2.5 J
SUBMIT

The distance covered by a ball in the first second of free fall is approximately 4.9 meters.

When an object falls freely under the influence of gravity, it experiences constant acceleration. In the case of Earth's gravity, the acceleration due to gravity is approximately 9.8 m/s². This means that the velocity of the falling object increases by 9.8 meters per second every second.

To determine the distance covered by the ball in the first second, we can use the equations of motion for uniformly accelerated motion.

The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is:

d = ut + (1/2)at²

In this case, the initial velocity is zero (as the ball starts from rest), the acceleration is 9.8 m/s², and we want to find the distance covered in the first second (t = 1 second).

Plugging in the values:

d = 0 * 1 + (1/2) * 9.8 * (1)^2

d = 0 + (1/2) * 9.8

d = 0 + 4.9

d = 4.9 meters

Therefore, the ball covers a distance of approximately 4.9 meters in the first second of free fall.

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Graphs can help to demonstrate the qualitative relationship between a set of data by identifying patterns among these data.

Graphs can allow readers to identify a lot of information among a set of data. The most common information to be evaluated through the graphs is the existence of patterns between the data.

For example, a line graph can show the change in a variable over time, allowing readers to see if there is a consistent increase or decrease.

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The reactive force when a fish swims through water is the force of the water on the fish. This is an example of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. The active force is the fish pushing against the water, so the reactive force would be the reverse, the equal force of the water pushing back on the fish.

(a) The work done by the donkey on the cart is 59,721.9 J.

(b) The work done by the force of gravity on the cart is -48,434.87 J.

(c) The work done on the cart by friction during this time is 11,315.12 J.

(a) The work done by the donkey on the cart is calculated as follows;

Wd = Fd cosθ

where;

Wd = 375 N x 163 m x cos(12.3)

Wd = 59,721.9 J

(b) The work done by the force of gravity on the cart is calculated as;

Wg =  Fg x d x cosθ

Where;

θ = 90⁰ + 4.03⁰ = 94.03⁰

Wg = (431 kg x 9.81 m/s²) x 163 m x cos (94.03)

Wg = -48,434.87 J

(c) The work done on the cart by friction during this time is calculated as;

Wf = Ff x d x cosθ

where;

Ff = μmg cosθ

Ff = 0.0165 x 431 kg x 9.81 x cos (4.03)

Ff = 69.59 N

Wf = 69.59 x 163 x cos (4.03)

Wf = 11,315.12 J

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The crate moves this time because the applied force of 150 N is greater than the maximum static friction force that opposes the motion of the crate. Static friction is a force that opposes the relative motion between two objects in contact. The maximum static friction force is determined by the normal force and the coefficient of static friction between the two surfaces in contact. The harder the surfaces are pushed together, the more force is needed to move them. When the applied force exceeds the maximum static friction force, the crate will start to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force.

The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C

To calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.

However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.

Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.

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There are 630,720,000 seconds in 20 years.

To calculate the number of seconds in 20 years, we need to consider the number of seconds in a minute, minutes in an hour, hours in a day, and days in a year.

1 minute consists of 60 seconds.

1 hour contains 60 minutes (60 minutes × 60 seconds = 3600 seconds).

1 day consists of 24 hours (24 hours × 3600 seconds = 86,400 seconds).

1 year typically consists of 365 days (365 days × 86,400 seconds = 31,536,000 seconds).

To find the number of seconds in 20 years, we multiply the number of seconds in one year by 20:

20 years × 31,536,000 seconds = 630,720,000 seconds.

Therefore, there are 630,720,000 seconds in 20 years.

This calculation assumes a non-leap year with 365 days.

If the 20 years span a leap year, the total number of seconds would be slightly higher, accounting for the extra day in the leap year.

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