What's the difference between mass and inertia in a tabular form

A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang

Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ = density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0

3 years ago

A 4.50-volt personal stereo uses 1950 joules of electrical energy in one hour. what is the electrical resistance of the personal

saveliy_v [14]

The stereo uses an energy of $E=1950 J$ in a time $t=1 h=3600 s$ , therefore the power of the stereo is given by
$P= \frac{E}{t}= \frac{1950 J}{3600 s}=0.54 W$

We also know that the power of an electrical device is related to its voltage, V, and its resistance, R, by the following equation
$P= \frac{V^2}{R}$
therefore, we can rearrange the equation to calculate the resistance of the stereo:
$R= \frac{V^2}{P}= \frac{(4.5 V)^2}{0.54 W}=37.5 \Omega$

6 0

4 years ago

Calculate the force of friction for a 5kg aluminum block being pulled with constant velocity (uniform motion) across a horizonta

Stolb23 [73]

The purpose of this lab is to determine whether the surface of an area would affect the coefficient of Friction. My classmates and I have learned a lot in this lab and that there could have been some errors in our lab because the strength of how a person pulls it might be a slight different than the normal force. I learned from this lab that the <span>surface area would have no effect on the coefficient of friction. </span>

4 0

3 years ago

I need help in my physics class and show me how it’s done

Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

$A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)$

Where | A | is the magnitude of the vector and $\alpha$ is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle $\alpha$ .

Vector A is in the 4th quadrant

So:

$A_x = 6\\\\A_y = -6.5$

So:

$|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846$

So:

$Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)$

$\alpha$ = -47.28 ° +360° = 313 °

$\alpha$ = 313 °

Option 4.

4 0

4 years ago

A rock is tossed straight up from ground level with a speed of 22 m/s. When it returns, it falls into a hole 9.0 m deep. a. What

Vera_Pavlovna [14]

Answer:

(a) $v_{f}=26.08m/s$

(b) $t=4.906seconds$

Explanation:

We will take ground level as origin and upward is positive direction

the givens are

$y_{i}=0\\y_{f}=-9.0m\\ v_{i}=22m/s$

Part (a)

To find the final velocity we use the kinematic equation

So

$(v_{f})^{2} =(v_{i})^{2} +2a(y_{f}-y_{i})\\(v_{f})^{2}=(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)\\v_{f}=\sqrt{(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)} \\v_{f}=26.08m/s$

Part (b)

To find time of rock trip until it touches the ground we will use simple kinematic equation or simple motion equation

$v_{f}=v_{i}+at\\t=\frac{v_{f}-v_{i}}{a}\\ t=\frac{(-26.08m/s)-(22m/s)}{-9.8m/s^{2} }\\t=4.906seconds$

Notice that we substituted vf with negative sign because its direction is downwards

5 0

3 years ago

What's the difference between mass and inertia in a tabular form​

What's the difference between mass and inertia in a tabular form