What is the difference in the speed of sound on a warm day versus on a cold day?

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:

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Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface cha

Ivan

Complete Question

The complete question is shown on the first uploaded image

Answer:

a it is always zero

b 0

c $\eta = -\epsilon _o E$

Explanation:ss

Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero

Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor

So We can mathematically define the charge density on the surface of the electric field as

∮ $E \cdot dA = \frac{-Q}{\epsilon _o}$

Where E is the electric field

$dA$ change in unit area

$-Q$ is the negative charge

$\epsilon _o$ is the permittivity of free space

So

$EA = \frac{-Q}{\epsilon _o }$

$\frac{Q}{A} = -\epsilon _o E$

$\eta = -\epsilon _o E$

Where $\eta$ is the charge density

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speed, volume, mass, temperature and power

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The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate

Sphinxa [80]

86.4×10^6 joule is energy does one house use during each 24 hr day.

20 MJ of light energy

Consumption of electricity is 1 kW.

The energy consumption lasts for 24 hours.

energy=power×time

energy=10^3×24×3600

energy=86.4×10^6 joule

Energy in physics is the ability to perform work. Different shapes, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc., may be assumed by it. Other examples of energy being transferred from one body to another include heat and work. Energy is always distributed after it has been transported in accordance with its type. Thus, heat transfer could result in thermal energy, whereas work could result in mechanical energy.

Motion is a trait shared by all forms of energy. For instance, if a body is moving, it has kinetic energy. Due to the object's design, which incorporates potential energy, a tensioned object, like a spring or bow, has the ability to move even when at rest.

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Which of the following is not a dwarf planet?

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The correct answer is (a.) Hydra. Hydra is not a dwarf planet, instead, it is the moon of the dwarf planet, Pluto. There are only four accepted dwarf planets by the International Astronomical Union which were the Haumea, Pluto, Eris, and Makemake.

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