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Zina [86]
3 years ago
9

A solution which has a solid solute is saturated at 25°C. It is then heated to 35°C. There is no change in the appearance of the

solution. What term would now be associated with this solution?
A.
unsaturated

B.
saturated

C.
oversaturated

D.
supersaturated
Physics
1 answer:
vampirchik [111]3 years ago
8 0

Answer:

C. oversaturated

Explanation:

Sana nakatulong

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A rocket is fired vertically upward. At the
Galina-37 [17]

Answer:

767.6 m/s

Explanation:

We consider momentum in both x and y components

Since momentum, p=mv where m denote mass while v is velocity then

In the y direction

3m*288=279m+my

y=585 m/s

In the x direction and taking right as positive then

3m*288=367m+mx

x=497 m/s

The resultant gives the velocity of the third fragment

Third=\sqrt{585^{2}+497^{2}}=767.6157893\approx 767.6 m/s

6 0
3 years ago
a hiker walks 11km due north from camp and then turns and walks 11 km due east. What is the approximate displacement of the hike
PIT_PIT [208]

Answer:

d=11\sqrt2\ km

Explanation:

Given that,

Displacement in North = 11 km

Displacement in East = 11 km

We need to find the approximate displacement of the hiker from the camp. It can be calculated by resultant of two displacement as follows :

d=\sqrt{11^2+11^2}\\\\= 11\sqrt2\ km

Hence, the approximate displacement of the hiker from the camp is 11\sqrt2\ km.

3 0
3 years ago
Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
andreyandreev [35.5K]
Disagree.
Fluoresce objects will only glow when put under actual Ultraviolet light. This is due to the molecules becoming excited by the ultraviolet radiation.


Microwaves give micro-waves that are present in another spectrum of wave length and will not be able to fluoresce the molecules. If it’s not “ultra violet “.... it’s not going to glow.
4 0
3 years ago
Read 2 more answers
I’m not sure how to do this
aleksley [76]
Do what?? i know how to do a lot of things to answer but there’s nothing to answer.‍♀️
6 0
4 years ago
1. A record with a radius of 0.3m spins in a clockwise circle with a centripetal
Hitman42 [59]

Solve for the linear/tangential speed:

<em>a</em> = <em>v</em>²/<em>r</em>

where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.

4.7 m/s² = <em>v</em>²/(0.3 m)

<em>v</em>² = (0.3 m) (4.7 m/s²)

<em>v</em> ≈ 3.96 m/s

For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is

(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s

Take the reciprocal of this to get the period:

1 / (7.46 rev/s) ≈ 0.134 s/rev

So it takes the record about 0.134 seconds to complete one revolution.

6 0
3 years ago
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