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Elina [12.6K]
3 years ago
11

you walk from the park to your friend's house, then back to your house. (a) What is the distance traveled? (b) What is your disp

lacement? If you walk from your house to the library, then back to your house, repeat (a) and (b).
Physics
1 answer:
777dan777 [17]3 years ago
3 0

Explanation:

The total distance in a path is called distance.

The shortest distance between two points is called displacement.

a) Here, the distance travelled between the park to your friend's house and back is

Distance between park to friends house + Distance from friend's house to your house.

b) Displacement would be the shortest distance between the park and your house.

a) Distance walked between your house to library and back is

Distance between your house and library + Distance between your house and library

b) Displacement would be zero (0) as the distance between you initial point and final point is zero. Here, the initial and final points are the same

You might be interested in
ASAP ASAP ASAP
mylen [45]

The experiments will involve two billiard balls of known masses, m₁ and m₂, and velocities u₁ and u₂. The two are allowed to collide and the velocities of the balls after the collision v₁ and v₂ are recorded.

The momentum before and after the collision is then calculated as follows:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

<h3>What is the statement of the law of conservation of momentum?</h3>

The law of the conservation of momentum states that the momentum before and after collision in a system of colliding bodies is conserved

The momentum of a body is calculated using the formula below:

Momentum = mass * velocity.

Hence, for the two billiard balls, the momentum before and after the collision is conserved.

Learn more about momentum at: brainly.com/question/1042017

#SPJ1

3 0
1 year ago
Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th
Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

\theta=30^{\circ}

So, the ball's horizontal velocity is

v_x = u cos \theta = (15)(cos 30)=13.0 m/s

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

y=u_yt+\frac{1}{2}at^2

where

u_y = u sin \theta = (15)(sin 30) = 7.5 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

d=v_x t =(13.0)(1.19)=15.5 m

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

d=98 m

while the horizontal velocity of the ball is

v_x = u cos \theta = (34)(cos 30)=29.4 m/s

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

y= h + u_y t + \frac{1}{2}at^2

where

h = 1.2 m is the initial height

u_y = u sin \theta = (34)(sin 30)=17.0 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting t = 3.33 s,

y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

u_y' = 7 m/s

So its position of the glove at time t' is

y'= h' + u_y' t' + \frac{1}{2}at'^2

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

t'' = t -t'

So we have two solutions:

t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
A book that weighs 19 Newtons sits on a table. With what force
iVinArrow [24]

Answer:

We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.

Explanation:

There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.

5 0
3 years ago
What is the objective of playing basagang palayok?the goal what is the goal of playing basang palayok ​
liubo4ka [24]

Answer:

Explanation:

At first glance, it looked like a group of kids-at-heart playing traditional Filipino games under the sun. But up close the games were played by public school teachers staging a unique protest to demand government to prioritize their welfare.

Some 100 members of the Teacher’ Dignity Coalition (TDC) from Metro Manila staged what they called “Palarong Pampista” at the Plaza Miranda in Manila yesterday to demand salary increases this year.

The protest is part of the “Protest de Mayo” launched by TDC earlier this month to call for a P10,000 across-the-board increase in salaries of government teachers and the scrapping of the performance-based bonus of the Aquino government which they described as “deceptive, unfair and divisive.”

During the protest, teachers played traditional fiesta games like Palo Sebo and Basagang Palayok “to demonstrate their sacrifices for the country despite the inadequate compensation from the government.”

TDC National Chairperson Benjo Basas said Palo Sebo and Basagang Palayok were symbols of how the government is treating public school teachers.

Palo Sebo, a popular fiesta game in which contestants climb a greased bamboo in order to get the money prize at the top, is the same as the government’s performance-based bonus or PBB, Basas said.

“It promises a cash prize, but teachers and employees need to fight and pull others down in order to be on top,” he explained.

The Basagang Palayok is also another popular fiesta game where the players’ objective is to hit the hanging pot with prizes. Basas said that in this game, “the player is blindfolded, and like the P10,000 salary increase demand, the prize is uncertain and the players must make all the effort and pass the obstacles.”

Earlier, Education Secretary Armin Luistro said that the Department of Education (DepEd) “recognizes the right of our teachers to raise their concerns publicly.”

“We will not object to any measure that will help public school teachers,” said Luistro. “As long as there is adequate funding, a raise in teachers’ salaries will be welcome,” he said.

However, while DepEd expressed support to the plight of public school teachers for salary increase, Luistro enjoined “certain groups not to take any action that greatly affects the delivery of basic services to our learners.”

TDC, a 30,000-strong group, has set several other unique mass actions until the resumption of classes in June 2 in all public elementary and secondary schools nationwide.

5 0
3 years ago
How far from the castle wall does the launched rock hit the ground?
lina2011 [118]
King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.
4 0
2 years ago
Read 2 more answers
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