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2 years ago

5

Marcia is given an incomplete chemical equation that includes the number of nitrogen atoms present in the products of the reacti

on. How can she conclude that the reaction started with three nitrogen atoms?

2 answers:

Semmy [17]2 years ago

8 0

<span>If the products contain 3 nitrogen atoms, then so did the reactants since overall mass is conserved in a chemical reaction. </span>

Sloan [31]2 years ago

5 0

Answer:There would have to be three nitrogen atoms in the products. The law of conservation of matter states that the amount of substance before a reaction occurs should be the same as the amount of substance after the reaction.

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Tom and his little sister are enjoying an afternoon at the ice rink. they playfully place their hands together and push against

Westkost [7]

Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".

So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span> $F_t = F_s$
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".

From Newton's second law, we also know
</span> $F=ma$
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span> $m_t a_t = m_s a_s$
<span>And find Tom's acceleration:
</span> $a_t = \frac{m_s}{m_t} a_s = \frac{15 kg}{61 kg} (2.1 m/s^2) =0.52 m/s^2$ <span>
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2 years ago

Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance as the two sys

baherus [9]

Answer:

a. Springs oscillate with the same frequency

Explanation:

As they both are in the same height at equilibrium, so

weight of ball must be balanced with spring force, that is

k×x=mg

k= stiffness constant of spring

x=distance stretched

g= acceleration due to gravity

so, we can write

k/m=g/x

as the g is a constant and they stretched to same distance x so the g/x term becomes constant and

$f\propto\sqrt{k/m}$

and k/m is same for both the springs so they will oscillate at the same frequency.

hence option a is correct.

3 0

2 years ago

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$

$\omega = 2\pi (2.5)$

$\omega = 5\pi rad/s$

The angular displacement is given as the form:

$\theta (t) = \theta_0 cos(\omega t)$

In the equlibrium we have to $t=0, \theta(t) = \theta_0$ and in the given position we have to

$\theta(t) = \theta_0 cos(5\pi t)$

Derived the expression we will have the equivalent to angular velocity

$\frac{d\theta}{dt} = 2.7rad/s$

Replacing,

$\theta_0(sin(5\pi t))5\pi = 2.7$

Finally

$\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°$

Therefore the maximum angular displacement is 9.848°

6 0

2 years ago

A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

<u>Answer:</u>

Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

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2 years ago

Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for th

Drupady [299]

<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>

<span> </span>

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2 years ago

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