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3 years ago

Do the same molecules have to be present on both sides of a chemical equation? why or why not?

1 answer:

Nadusha1986 [10]3 years ago

7 0

Yes, the same molecules have to be present on both sides of a chemical equation. This is because the equation has to be able to be balanced, and without the same amount of each molecule on each side of the equation, it could not be balanced.

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Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

4 years ago

In a golf ball game, a person hits the golf ball with a club. The club is in contact with the ball, which is initially at rest,

Anon25 [30]

Answer:

F=4040.81 N

Explanation:

Given that

Time ,t= 2.45 ms

Mass ,m= 0.055 kg

v= 1.8 x 10² m/s = 180 m/s

We know that rate of change in the linear momentum is known as force.

Momentum P = m v

$F=\dfrac{dP}{dt}$

Therefore force F

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{0.055\times 180}{2.45\times 10^{-3}}\ N$

F=4040.81 N

Therefore force on the ball will be 4040.81 N

6 0

3 years ago

Un semáforo de 80N cuelga del punto medio de un cable de 30m tendido entre 2 postes. Halle la tensión en cada segmento del cable

olganol [36]

Answer:

T1 = T2 = 602.33 N

Explanation:

I attached an image of the forces diagram below.

The x component of the forces for this cases is:

$\Sigma F_x=T_1cos\theta-T_2cos\theta=0\\\\T_1cos\theta=T_2cos\theta\\\\T_1=T_2$

And the y components:

$\Sigma F_y=T_1sin\theta+T_2sin\theta-W=0\\\\T_1=T_2=T\\\\\Sigma F_y=2Tsin\theta=W\\\\T=\frac{W}{2sin\theta}$

The angle is calculated by using the information about the length of the cable and the vertical distance of the traffic light:

$\theta=tan^{-1}(\frac{1}{15})=3.81\°$

Thus, you obtain:

$T=\frac{80N}{2(sin3.81\°)}=601.33N$

hence, the tension in both segments of the cable is 602.33 N

3 0

4 years ago

Plz help me it is improtant

marishachu [46]

I think it is b cause I don’t think you do that

7 0

3 years ago

Which of the following might be the material list for an experiment?

Alisiya [41]

Answer:

Explanation:

8 0

4 years ago