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34kurt
3 years ago
10

Isla made ice by putting water in the freezer. Freezing is an example of (2 points)

Physics
2 answers:
alexgriva [62]3 years ago
7 0

Answer:

a physical change

Explanation:

because water as it is in its original state a liquid is frozen therefore it changed its physical look

SSSSS [86.1K]3 years ago
6 0

Answer:

a physical change

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A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi
zubka84 [21]
The answer is A. Polarized in a vertical plane

If positioned correctly, a polarized lenses can block all reflected light from horizontal surface such as road
8 0
3 years ago
Read 2 more answers
A cyclist maintains a constant velocity of 6.1 m/s headed away from point A. At some initial time, the cyclist is 242 m from poi
KengaRu [80]

Answer:

553.1m

Explanation:

When an object moves at constant velocity we can express this movement like V=x/t, where V is the velocity, x is the displacement and t is the time spent on it.

In that way, the expression x=V.t give us the displacement from t=0s until t=51s, but we have to sum the initial distance from the point A.

x=242m +V.t = 242m + (6.1m/s x 51s) = 553.1m

7 0
3 years ago
What's the velocity for each trial? <br> And what's the average velocity?
adelina 88 [10]
Since each time trial is the same the average will be the direct answer, and the formula for velocity is distance divided by time, therefore it will come out to, 1.92307692. Whatever your teacher what the rounding process to be will vary but the straight up answer is there.
5 0
3 years ago
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