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sergiy2304 [10]
3 years ago
14

Low-viscosity lava a. is most often a cool-temperature lava. b. could logically build a composite volcano. c. has low silica con

tent. d. indicates an area that has high potential for explosive eruptions.
Physics
1 answer:
OverLord2011 [107]3 years ago
8 0

<u>Full question:</u>

Identify the False statement. Lava of low viscosity

a. is most often a cool-temperature lava.

b. could logically build a composite volcano.

c. has low silica content.

d. indicates an area that has high potential for explosive eruptions.

<u>Answer:</u>

is most often a cool-temperature lava is the false statement about  Lava of low viscosity

<u>Explanation:</u>

When lava begets low viscosity, it can move quite smoothly across extended lengths. This produces the perfect outpourings of lava, byways, plashes, and sprays. You can likewise notice globules of lava-filled amidst volcanic gasses that burble and rise on the facade of the lava.

And extra time, volcanoes produced from low lava viscosity are extensive and have a depthless incline certain are perceived as shield volcanoes. Anywhere a volcano provides low viscosity, runny, lava it flattens faraway from the origin producing a volcano with moderate inclines.

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2. CaCl2 (s) + 2H20 ---&gt; Ca(OH)2 (aq) + 2HCl (g) + heat<br> Is it an endothermic reaction?
antoniya [11.8K]

Answer:

no

Explanation:

Calcium chloride is a chemical compound made up of calcium ions and chlorine ions. ... Mixing calcium chloride with water is an exothermic reaction, which means that the combination of the two substances releases heat. Thus, when you add calcium chloride to water, the solution heats.

3 0
3 years ago
the pilot of a new stealth helicopter which has a mass of 15000 kg and was traveling 180 m / s accelerated to 250 m / s in six s
MariettaO [177]

Answer:

<em>pf = 3750000 kg.m/s</em>

<em>F = 175000 N</em>

Explanation:

<u>Constant Acceleration</u>

An object moves with constant acceleration if its velocity changes uniformly in time.

If we call vo to the initial speed, vf the final speed, and t the time, then the acceleration is calculated as:

\displaystyle a=\frac{v_f-v_o}{t}

The new stealth helicopter was traveling at vo=180m/s and changed to vf=250 m/s in t=6 seconds, thus the acceleration was:

\displaystyle a=\frac{250-180}{6}

a=11.67~m/s^2

<em>Please note this number is shown rounded to the nearest hundredth, but it was stored in the calculator's memory with full precision. This fact affects the next calculation as will be noted below.</em>

The acceleration must be produced for some net force that can be calculated by:

F = m.a

Where m=15000 Kg is the mass of the helicopter. Thus:

F = 15000 * 11.67

F = 175000 N

<em>The above product would lead to the inaccurate result of 175050 if the acceleration would have not used with its full representation in the calculator's memory.</em>

The momentum of an object of mass m and velocity v is:

p = m. v

The final momentum of the helicopter is:

pf = m.vf = 15000*250 m/s

pf = 3750000 kg.m/s

<u>Note: When performing calculations over intermediate results, it's important to keep them as accurate as possible to preserve the accuracy of the final result.</u>

6 0
3 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
3 years ago
The liquid in the cup is brown" is an example of an ..............
erastova [34]
Eaither D or A bit I am leaning more towards D
3 0
3 years ago
Read 2 more answers
The small ball of mass m = 0.5 kg is attached to point A via string and is moving at constant speed in a horizontal circle of ra
babymother [125]

Answer:

d = 2.45 meters

Explanation:

Mass of the ball, m = 0.5 kg

Radius of the circle, r = 0.16 m

The angular speed of the ball around the circle is, \omega=2\ rad/s

The attached figure shows the whole scenario. Let F_t is the force acting on the ball in tangential direction. The forces will balanced each other at equilibrium.

In horizontal direction,

T\ sin\theta=F_t=mr\omega^2................(1)

In vertical direction,

T\ cos\theta=mg...............(2)

From equation (1) and (2) :

tan\theta=\dfrac{r\omega^2}{g}

Also,

tan\theta=\dfrac{r}{d}

d=\dfrac{g}{\omega^2}d=\dfrac{9.8}{2^2}

d = 2.45 meters

So, the value of d is 2.45 meters. Hence, this is the required solution.  

3 0
4 years ago
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