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luda_lava [24]
2 years ago
11

A mass of (200 g) of hot water at (75.0°C) is mixed with cold water of mass M at (5.0°C). The final temperature of the mixture i

s (25.0°C). What is the mass of the cold water (M)?
Physics
1 answer:
MArishka [77]2 years ago
3 0

Answer:

500 g

Explanation:

Specific heat of water = 1 j/g-c

Heat given up by hot water = 200(75-25)(1) = 10 000 j

this is the heat GAINED by the cold water

10000 j =  x ( 25 -5)(1)  

          x =500g

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Mass (kg) 4.0
densk [106]

Answer:

25 m/s

Explanation:

First of all, we can find the acceleration the object by using Newton's second law of motion:

F=ma

where

F = 20.0 N is the net force applied on the object

m = 4.0 kg is the mass of the object

a is its acceleration

Solving for a, we find

a=\frac{F}{m}=\frac{20}{4}=5.0 m/s^2

Now we know that the motion of the object is a uniformly accelerated motion, so we can find its final velocity by using the following suvat equation:

v=u+at

where

v is the final velocity

u = 0 is the initial velocity

a=5.0 m/s^2 is the acceleration

t = 5 s is the time

By substituting,

v=0+(5.0)(5)=25 m/s

8 0
3 years ago
How should ready-to-eat food be stored after it has cooled?
snow_lady [41]

Answer:

The answer is most likely C.

Ready-to-eat foods are stored at the top of the fridge, away from raw foods so that harmful bacteria cannot transfer from the raw food to the cooked food. Raw meat, poultry and fish in sealed containers to stop them touching or dripping onto other foods.

5 0
3 years ago
¿Cuál es la intensidad del campo eléctrico a una distancia de 2 m de una carga de -12μC? *
ahrayia [7]

Answer:

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3 years ago
It is important that your muscles are very warm when doing this type of stretching.
MaRussiya [10]

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3 0
3 years ago
Read 2 more answers
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

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a'=4\times a

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
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