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2 years ago

5

What determines whether the equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler

or warmer water?

1 answer:

larisa [96]2 years ago

6 0

Answer:

Explanation:

It is determined by the amount of cold or hot water in the mixture to know if it will be closer to been hot or cold.

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The international governing body for badminton is the

vekshin1

<span>The Badminton World Federation</span>

4 0

3 years ago

Read 2 more answers

I appreciate any help I get on this, thank you in advanced :)

Reika [66]

The picture isn’t clear so I can’t read the dimensions of the box but I can try my best to guide u through the question.

For part a u need to find the volume of the box as that will equal the volume of sand that can be filled inside.
For this u multiply the height, width and length of the box.

For part b the mass of sand alone will be
=Mass of box + sand - Mass of empty box
=216 - 40
=176 grams

For part c the density of sand can be calculated by the formula
Density= Mass/Volume
So the mass (176g) / volume from part a

For part d u need to know that something will float if it has a lower density than what it is floating in. If the final density of sand that was found in part c is less than the density of gold (19.3 g/cm^3) it will float. Otherwise it will sink.

Hope this helped!

3 0

2 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

2 years ago

PLEASE HLEP I WILL GIVE BRAINLIEST TO CORRECT ANSWER!!!!!!

Hoochie [10]

Answer:

75 km/h

Explanation:

Speed = Distance divided by Time.

The train went a distance of 300 km in 4 hours. Therefore, the speed of the train is: 300 divided by 4 = 75 km/h

8 0

2 years ago

Read 2 more answers

A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

$a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2$

the angular speed is given by:

$\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s$

to calculate how many radians have the wheel turned we need the apply the following formula:

$\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad$

the distance is given by:

$d=\theta*r$

$d=381rad*(35.5*10^{-2}m)\\d=135m$

4 0

3 years ago