PHYSICS HIGH SCHOOL

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?

Answers

Answer 1

Answer:

The speed of the ball just before it strikes the ground is equal to 46.55 m/s.

Given the following data:

Horizontal distance = 65 meters

Height of building = 0.10 km = 100 meters

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$ .

To determine the speed of the ball just before it strikes the ground:

First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.

$H = (1)/(2) gt^2\n\n100 = (1)/(2) * 9.8 * t^2\n\n200 = 9.8t^2\n\nt^2 = (200)/(9.8) \n\nt^2=20.41\n\nt=√(20.41)$

Time, t = 4.52 seconds

Next, we would find the horizontal velocity:

$Horizontal\;velocity = (horizontal\;distance)/(time) \n\nHorizontal\;velocity = (65)/(4.52)$

Horizontal velocity, V1 = 14.38 m/s

Also, we would find the velocity of the ball in the horizontal direction:

$V_2^2 = U^2 + 2aS\n\nV_2^2 = 0^2 + 2(9.8)(100)\n\nV_2^2 = 1960\n\nV_2 = √(1960) \n\nV_2 = 44.27 \;m/s$

Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:

$V = √(V_1^2 + V_2^2) \n\nV = √(14.38^2 + 44.27^2)\n\nV = √(206.7844‬ + 1959.8329‬)\n\nV =√(2166.6173)$

Speed, V = 46.55 m/s

Answer 2

Answer:

Answer:

v = 46.55 m/s

Explanation:

It is given that,

A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m

The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m

At maximum height, velocity of the ball is 0. So, using the equation of motion as :

$d=ut+(1)/(2)at^2$

Here, a = g

$100=0+(1)/(2)* 9.8t^2$

$t=4.51\ s$

Let $v_x$ is the horizontal velocity of the ball. It is calculated as :

$v_x=(65\ m)/(4.51\ s)=14.41\ m/s$

Let $v_y$ is the final speed of the ball in y direction. It can be calculated as :

$v_y^2+u_y^2=2as$

$u_y=0$

$v_y^2=2gd$

$v_y^2=2* 9.8* 100$

$v_y=44.27\ m/s$

Let v is the speed of the ball just before it strikes the ground. It is given by :

$v=√(v_x^2+v_y^2)$

$v=√(14.41^2+44.27^2)$

v = 46.55 m/s

So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.

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A horizontal 20.-newton force is applied to a 5.0-kilogram box to push it across a rough,horizontal floor at a constant velocity of 3.0 meters per second to the right.

Calculate the coefficient of kinetic friction between the box and the floor. [Show all work, including
the equation and substitution with units]

Answers

The formula to work out the coefficient of dynamic (kinetic) friction is F=uR where F is the force applied, u is the coefficient of dynamic friction, and R is the reaction force. This is based on the principle the object is moving at a constant velocity which is the case in this question.

Rearranging this we get u=F/R
Substituting in the values we get u=20/50=2/5=0.4

Therefore the coefficient of kinetic friction between the box and the floor is 0.4

The coefficient of kinetic friction between the box and the floor will be $\boxed{0.4}$ .

Further Explanation:

Given:

The mass of the box is $5\,{\text{kg}}$ .

The force applied on the box is $20\,{\text{N}}$ .

The constant velocity at which the box moves is $3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

Concept:

The box is pushed on the rough floor and due to this; the box will experience a friction force acting in the direction opposite to the motion of the box.

Since the box moves with a constant velocity of $3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ , the acceleration of the box will be considered as zero.

From Newton’s second law of motion, as the acceleration of the box is zero, the net force acting on the box is zero.

$\boxed{{F_(net)}=m* a}$

The free-body diagram of the box moving on the rough floor is as shown in figure attached below.

From the above explanation, in order to make the net force zero, the friction acting on the box should be equal to the force applied on the box.

${F_(friction)}={F_(applied)}$

The friction force acting on the bxlock is:

$\begin{aligned}{F_(friction)}&=\mu N\n&=\mu mg\n\end{aligned}$

Substitute the value of friction force in the above expression.

$\begin{aligned}\mu mg&={F_(applied)}\n\mu\left({5*10}\right)&=20\n\mu &=\frac{{20}}{{50}}\n&=0.4\n\end{aligned}$

Thus, the coefficient of kinetic friction between the box and the floor will be $\boxed{0.4}$ .

Learn More:

1. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N brainly.com/question/7031524

2. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right). Be sure friction is turned off brainly.com/question/4033012

3. Which of the following is not a component of a lever brainly.com/question/1073452

Answer Details:

Grade: High School

Subject: Physics

Chapter: Friction

Keywords:

Horizontal 20N force, 5 kg box, push, a rough horizontal floor, constant velocity, friction force, coefficient of kinetic friction, box and the floor.

In a nuclear reactor, the moderator is: usually made of graphite placed between the fuel rods used to slow down neutrons all of the above none of the above

Answers

The correct answer is all of the above. In a nuclear reactor, the moderator is usually made of graphite, placed between the fuel rods, and used to slow down neutrons. The nuclear moderator is a medium in which reduces the speed of the fast neutrons.

The answer is D. All the above

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Answers

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Answers

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What is the minimum total energy released whenan electron and its antiparticle (positron)
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(1) 1.64 × 10^–13 J (3) 5.47 × 10^–22 J
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Answers

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Explanation:

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The minimum total energy released when an electron and its antiparticle (positron) annihilate each other is 1.64 x 10⁻¹³ J.

What is the minimum total energy released?

The minimum total energy released when an electron and its antiparticle (positron) annihilate each other is calculated as follows;

E = Δmc²

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m is the mass of the two electrons
c is speed of light

E = (2 x 9.11 x 10⁻³¹) x (3 x 10⁸)²

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Why did most scientist reject wegeners theory for nearly a half century

Answers

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