A ball is allowed to fall freely from certain height it covers a distance of 1st sec equal to?

Answers

Answer 1

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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Related Questions

Part C
Increase the value of the Applied Force to 150 N. Pause the simulation and observe the magnitudes and directions of the applied force, the friction
force, and the sum of forces. Why do you think the crate moves this time?

Answers

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.

how many seconds makes 20 years (show all workings)​

Answers

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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Refer to the picture!

Answers

(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.

What is the work done by the donkey on the cart?

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

Wd = Fd cosθ

where;

F is the applied force by the donkeyd is the displacementθ is the angle of inclination

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;

Fg is the force of gravityd is the displacementθ is the angle between the force of gravity and displacement

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 is the force of friction;

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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What is the reactive force when a fish swims through water

Answers

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 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

Answers

The kinetic energy of the book after it is slids a distance of 2 meters will be 10 Joules.

How to determine the kinetic energy of an object?

The work-energy theorem states that "the work done on an object is the change in its kinetic energy".

Hence;

Kinetic energy = work done

Note that: work-done is expressed as:

Work done = f × d

Where f is force applied and d is distance traveled.

Given that:

Force applied f = 5 newton

Distance d = 2 meters

Work done = ?

Plug these values into the above formula and solve for the workdone.

Work done = f × d

Work done = 5N × 2m

Work done = 10Nm

Work done = 10 Joules

Therefore, the kinetic energy is 10 Joules.

Option B) 10 J is the correct answer.

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Select the correct answer.
A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward point A. What are the total
distance and total displacement of the boat?
O A.
OB.
O C.
O D.
The total distance is 105 kilometers and the total displacement is 45 kilometers east.
The total distance is 60 kilometers and the total displacement is 60 kilometers east.
The total distance is 105 kilometers and the total displacement is 15 kilometers east.
The total distance is 60 kilometers and the total displacement is 45 kilometers east.

Answers

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

How to solve for the  total distance

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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PLEASE HELP! Thank you!
Chloe and Sarah are driving bumper cars. Chloe, who is traveling west at 3.9 m/s, is behind Sarah, who is traveling west at 1.6 m/s. The total mass of Chloe’s car is 163 kg, and total mass of Sarah’s car is 179 kg. Immediately after Chloe collides with Sarah, Chloe’s velocity reduces to 0.95 m/s west. What is Sarah’s velocity immediately after the collision?

A. 5.2 m/s
B. 4.0 m/s
C. 4.3 m/s
D. 4.6 m/s

Answers

Sarah’s velocity immediately after the collision is 4.3 m/s west.

option C is the correct answer.

What is Sarah’s velocity immediately after the collision?

Sarah’s velocity immediately after the collision is calculated by applying the principle of conservation of linear momentum as follows;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

m₁ is the mass of Sarah's carm₂ is the mass of Chloe's caru₁ is the initial velocity of Sarahu₂ is the initial velocity of Chloev₁ is the final velocity of Sarahv₂ is the final velocity of Chloe

Sarah’s velocity immediately after the collision is calculated as;

179 (1.6) + 163(3.9) = 179v₁ + 163(0.9)

922.1 = 179v₁ + 146.7

179v₁ = 775.4

v₁ = 4.3 m/s

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How can graphs help demonstrate the qualitative relationship that may exist in a set of data to readers?

Answers

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

What other information can the data show?Comparison between groups.Proportional relationships.Variation and dispersion.

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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