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Harlamova29_29 [7]

3 years ago

7

In a video game a flying coconut moves at a cost of the last few 20 m/s a positive direction the coconut he stops going to move

in the opposite direction with the cost of the last we have 10 m/s what is the change in velocity of the coconut

1 answer:

Liula [17]3 years ago

6 0

Answer:

30 m/s

Explanation:

30 m/s I think.

I am too lazy to explain

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1. A concave mirror has a focal length of 1.50 meters. What is the radius of curvature of the mirror? An object is placed 4.00 m

matrenka [14]

1) 3.0 m

2) 2.40 m

Explanation:

1)

A concave mirror is a reflecting surface that causes the reflection of the rays of light coming to the mirror, producing an image of the object facing the mirror.

There are two types of mirror:

- Concave mirror: this is curved inward - as a result, the rays of light coming from the object are reflected back into a single point, called focal point

- Convex mirror: this is curved outward - as a result, the rays of light coming from the object are reflected back into diverging direction, not into a single point

For a curved mirror, the radius of curvature is twice the focal length:

$R=2f$

Where

R is the radius of curvature

f is the focal length

In this problem,

f = 1.50 m

So, the radius of curvature is

$R=2(1.50)=3.0 m$

2)

The distance of the image from the mirror can be found by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

IN this problem we have:

f = 1.50 m is the focal length

p = 4.00 m is the distance of the object from the mirror

Solving for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.50}-\frac{1}{4.00}=0.416\\q=\frac{1}{0.416}=2.40 m$

8 0

3 years ago

A football player runs from his own goal line to the opposing team's goal line, returning to his forty-yard line, all in 22.4 s.

Bumek [7]

I assume L=120 yards as the length of the football field.

1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
$S=120y+(120-40)y=120y+80y=200 y$
And the time taken is t=22.4 s, so the average speed of the player is
$v= \frac{S}{t}= \frac{200 y}{22.4 s}=8.93 y/s$

2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
$S=120 y-80 y=40 y$
and so, the average velocity is
$v= \frac{S}{t}= \frac{40 y}{22.4 s}= 1.79 y/s$

6 0

3 years ago

Calculate the mass of 1.35 moles of sodium chloride (NaCl).

Harlamova29_29 [7]

Answer:d

Explanation:

4 0

3 years ago

what is the period of a sound wave whose wavelength is 20.0m? use values from the book and show all of your work

Lera25 [3.4K]

The wavelength of a sound wave is related to its frequency by the relationship:
$f= \frac{v}{\lambda}$
where
f is the frequency
v is the speed of the wave
$\lambda$ is the wavelength

The wave in our problem has wavelength of $\lambda=20.0 m$ and speed of $v=343 m/s$ (this is the speed of sound in air), therefore its frequency is
$f= \frac{343 m/s}{20.0 m}=17.15 Hz$

And the period of the wave is equal to the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{17.15 Hz}=0.058 s$

5 0

4 years ago

A spherical shell has an outside radius of 2.60 cm and an inside radius of a. The shell wall has uniform thickness and is made o

lbvjy [14]

Answer:

The answer is below

Explanation:

a) The volume of a sphere is:

Volume = (4/3)πr³; where r is the radius of the shell.

Given the outside radius of 2.60 cm and inner radius of a cm, the volume of the spherical shell is:

Volume of spherical shell = $\frac{4}{3} \pi (2.6^3-a^3)$ cm³

But Density = mass / volume; Mass = density * volume.

Therefore, mass of spherical shell = density * volume

mass of spherical shell = $4.70\ g/cm^3$ * $\frac{4}{3} \pi (2.6^3-a^3)$ cm³

Mass of liquid = volume of inner shell * density of liquid

Mass of liquid = $\frac{4}{3} \pi a^3\ cm^3*1.23\ g/cm^3$

Total mass of sphere including contents = mass of spherical shell + mass of liquid

Total mass of sphere including contents (M) = $4.70\ g/cm^3$ * $\frac{4}{3} \pi (2.6^3-a^3)\ cm^3$ + $\frac{4}{3} \pi a^3\ cm^3*1.23\ g/cm^3$ =

Total mass of sphere including contents (M) = (346 - 14.5a³) grams

b) The mass is maximum when the value of a = 0

M = 346 - 14.5a³

M' = 43.5a² = 0

43.5a² = 0

a = 0

4 0

3 years ago