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aivan3 [116]
3 years ago
5

A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th

e solution if the density is 1.12 g/mL.
Physics
1 answer:
son4ous [18]3 years ago
7 0

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass =volume\times density

Using the formula

Mass of solution=1.12\times 473=529.76 g

Mass % of KBr=\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100

Mass % of KBr=\frac{49.3}{529.76}\times 100

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

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Answer:

Explanation:

heat lost by water will be used to increase the temperature of  ice

heat gained by ice

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1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt  where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186  

8372

heat lost = heat gained

1 x 2090 x t  = 8372

t = 4°C

There will be a rise of  4 degree in the temperature of ice.  

 

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3 years ago
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I think the correct answer is is D.
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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee
emmainna [20.7K]

a) 0.0028 rad/s

b) 23.68 m/s^2

c) 0 m/s^2

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

\omega = \frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

\omega = \frac{v}{r} (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

v=29,960 km/h is the linear speed, converted into m/s,

v=8322 m/s

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

a_r=\omega^2 r

where

\omega is the angular speed

r is the radius of the orbit

For the spaceship in the problem, we have

\omega=0.0028 rad/s is the angular speed

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

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v=8322 m/s

Therefore, its linear speed is not changing, so the change in linear speed is zero:

\Delta v=0

And therefore, the tangential acceleration is zero as well:

a_t=\frac{0}{\Delta t}=0 m/s^2

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