What has to happen to an uncharged aluminum atom so that it will bond with oxygen?

If you know the power rating of an appliance and the current the appliance uses, you can calculate the voltage of the line it is

Helga [31]

If their are choices you should list them. The givens are pretty specific though.

W = E * I is the answer
E (voltage) = W/I is a more refined answer. <<< answer.

7 0

2 years ago

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A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig

inna [77]

Answer:

$\mu_s=\frac{1}{3}\tan \theta$

Explanation:

Let the minimum coefficient of static friction be $\mu_s$ .

Given:

Mass of the cylinder = $M$

Radius of the cylinder = $R$

Length of the cylinder = $L$

Angle of inclination = $\theta$

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

$\tau =I\alpha$

Now, angular acceleration is given as:

$\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}$

Also, moment of inertia for a cylinder is given as:

$I=\frac{MR^2}{2}$

Therefore, the torque acting on the cylinder can be rewritten as:

$\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1$

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are $mg\sin \theta\ and\ f$ . The net force acting along the incline is given as:

$F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2$

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, $N=Mg\cos \theta$

Plugging in $N=Mg\cos \theta$ in equation (2), we get

$F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3$

Now, as per Newton's second law,

$F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4$

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

$\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)$

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

$\mu_s=\frac{1}{3}\tan \theta$

3 0

2 years ago

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PART ONE

stira [4]

Answer:

3.64×10⁸ m

3.34×10⁻³ m/s²

Explanation:

Let's define some variables:

M₁ = mass of the Earth

r₁ = r = distance from the Earth's center

M₂ = mass of the moon

r₂ = d − r = distance from the moon's center

d = distance between the Earth and the moon

When the gravitational fields become equal:

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

M₁ / r² = M₂ / (d − r)²

M₁ / r² = M₂ / (d² − 2dr + r²)

M₁ (d² − 2dr + r²) = M₂ r²

M₁d² − 2dM₁ r + M₁ r² = M₂ r²

M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0

d² − 2d r + (1 − M₂/M₁) r² = 0

Solving with quadratic formula:

r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)

When we plug in the values, we get:

r = 3.64×10⁸ m

If the moon wasn't there, the acceleration due to Earth's gravity would be:

g = GM / r²

g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²

g = 3.34×10⁻³ m/s²

4 0

2 years ago

What is a measure of the amount of matter in an object? Question 2 options: Force Inertia Mass Acceleration

fredd [130]

The answer is C.) mass is the matter of an object

6 0

3 years ago

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Two plane mirrors are separated by 120°, as the drawing illustrates. If a ray strikes mirror M1 at a θ1 = 64° angle of incidence

tino4ka555 [31]

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

= 90° - 64°

= 26°

β + γ + 120° = 180°

γ = 180° - 120° - β

γ = 180° - 120° - 26°

γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

δ = 56°

According to the law of reflection,

angle of incidence = angle of reflection

θ2 = δ = 56°

Therefore, angle θ2 at which the light leaves mirror 2 is 56°

8 0

2 years ago