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3 years ago

How do I calculate density?

1 answer:

4 0

The Density Calculator uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). The calculator can use any two of the values to calculate the third. Density is defined as mass per unit volume.

question answered by

(jacemorris04)

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Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

lbvjy [14]

Answer:

Electric flux;

Φ = 30.095 × 10⁴ N.m²/C

Explanation:

We are given;

Charge on plate; q = 17 µC = 17 × 10^(-6) C

Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²

Angle between the normal of the area and electric field; θ = 4°

Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m

Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²

The charge density on the plate is given by the formula;

σ = q/A_p

Thus;

σ = (17 × 10^(-6))/(180 × 10^(-4))

σ = 0.944 × 10^(-3) C/m²

Also, the electric field is given by the formula;

E = σ/ε_o

E = (0.944 × 10^(-3))/(8.85 × 10^(-12))

E = 1.067 × 10^(8) N/C

Now, the formula for electric flux for uniform electric field is given as;

Φ = EAcos θ

Where A = πr² = π × 0.03² = 9π × 10^(-4) m²

Thus;

Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4

Φ = 30.095 × 10⁴ N.m²/C

3 0

3 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

3 years ago

vfiekz [6]

Answer:

B. fjords

explanation :

Fjords were created by glaciers. In the Earth's last ice age, glaciers covered just about everything. Glaciers move very slowly over time, and can greatly alter the landscape once they have moved through an area. This process is called glaciation.The fjords are one of the glaciers that existed in the past. They are one of the many glacial relief forms that can give us a insight into the size and power of the glaciers.

5 0

3 years ago

A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the cloth

nadya68 [22]

Answer:

The tension on the clotheslines is $T = 8.83 \ N$

Explanation:

The diagram illustrating this question is shown on the first uploaded image

From the question we are told that

The distance between the two poles is $d = 12 \ m$

The mass tie to the middle of the clotheslines $m = 1 \ kg$

The length at which the clotheslines sags is $l = 4 \ m$

Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as

$W = mg$

let the angle which the tension on the clotheslines makes with the horizontal be $\theta$ which mathematically evaluated using the SOHCAHTOA as follows

$Tan \theta = \frac{ 4}{6}$

=> $\theta = tan^{-1}[\frac{4}{6} ]$

=> $\theta = 33.70^o$

So the vertical component of this tension is mathematically represented a

$T_y = 2* Tsin \theta$

Now at equilibrium the net horizontal force is zero which implies that

$T_y - mg = 0$

=> $T sin \theta - mg = 0$

substituting values

$T = \frac{m*g}{sin (\theta )}$

substituting values

$T = \frac{1 *9.8}{2 * sin (33.70 )}$

$T = 8.83 \ N$

6 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago