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2 years ago

9

In the equation Ca(s) + 2H2O(l) -->Ca(OH)2(aq) + H2(g) list each element used as a reactant, and tell how many atoms of each

are needed for this reaction.

1 answer:

Mekhanik [1.2K]2 years ago

3 0

The reactants are <em>Calcium</em>, <em>Hydrogen</em> and <em>Oxygen</em>.

We need 1 atom of <em>Calcium</em>, 4 atoms of <em>Hydrogen</em> and 2 atoms of <em>Oxygen</em>.

Hope this helps.

r3t40

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A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.

Shalnov [3]

Answer:

Explanation:

During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s

v = u + at

1.2 = 0 + .8 a

a = 1.2 / .8

= 1.5 m /s²

During the acceleration in upward direction , let reaction force of ground on man be R .

Net force on man = R - mg

Applying Newton's 2 nd law

R - mg = ma

R = m ( g + a )

= 72 ( 9.8 + 1.5 )

= 813.6 N .

This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .

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2 years ago

Extrusive rocks forms beneath earth's surface true or false

Verizon [17]

The answer is false your welcome

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3 years ago

A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

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3 years ago

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

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3 years ago

Read 2 more answers

In a crossing situation, which vessel is required to maintain its course and speed?

makvit [3.9K]

Both in the domestic and international guidelines tell that when two power-driven vessels are crossing so as to contain risk of collision, the vessel which has the other on her starboard side (the give-way vessel) must keep out of the way.

If you are the give-way vessel, it is your responsibility to avoid a collision. Normally, this means you must change speed or direction to cross behind the other vessel which is the stand-on vessel.

At evening, when you perceive a red light crossing right-to-left in front of you, you need to change your course. But if you perceive a green light crossing from left-to-right, you are the stand-on vessel, and should maintain course and speed.

The leading situations of collision risk are meeting head-on, overtaking, and crossing. When one of two vessels is to keep out of the way (give-way vessel), the other, the stand-on vessel, must uphold course and speed.

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2 years ago