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Kisachek [45]
3 years ago
6

List three signs that could make you think that a chemical reaction was taking place.

Physics
1 answer:
algol133 years ago
5 0
1. Release of gases
2. Bubbling 
3. And change in color- this one can also be for a physical change.

Hope this helps.
 
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Examine figure 26-1 which area is the convection zone<br><br>A. A<br>B. B<br>C. C<br>D. D
kykrilka [37]
I think the answer is B
4 0
3 years ago
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Bruno the bat flies at a speed of 0.5 m/s in circle of radius 1 m. What is his acceleration?
beks73 [17]

Answer:

Acceleration is 0.25m/s^2

Explanation:

Given the following :

Speed = 0.5m/s

Radius(r) of circle = 1m

Acceleration round a circular path is given as :

a = v^2 / r

Where

a = acceleration of the body

v = speed / velocity

r = radius

Therefore,

a = v^2 / r

a = (0.5)^2 / 1

a = 0.25m/s^2

7 0
3 years ago
What two forces act on a falling object ?
goldenfox [79]

Answer:

The two forces acting on the object are weight due to gravity pulling the object towards earth, and drag resisting this motion. When the object is first released, drag is small as velocity is low, so the resultant force is down. This means the object accelerates towards earth.

5 0
3 years ago
One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied
Alexandra [31]

Answer: 1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

8 0
3 years ago
you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia
Ivanshal [37]

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m

So its area must be

A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s

7 0
3 years ago
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